The motion of a particle with position (x,y) as t varies in the given interval is given by the following.x=3+sin(t), y=5+6cos(t), π/2 ≤ t ≤ 2πThe motion of the particle takes place on an ellipse centered at (x,y) = (3, 5).As we know that the general equation of an ellipse is given by ((x-h)²/a²)+((y-k)²/b²)=1Where h and k are the center of the ellipse and a and b are the lengths of the semi-major axis and semi-minor axis respectively.
Comparing the given equation with the general equation of the ellipse, we have: Centre of ellipse = (h,k) = (3, 5)Length of semi-major axis a = 1Length of semi-minor axis b = 6Hence, the particle moves on an ellipse with center (3, 5), semi-major axis of length 1, and semi-minor axis of length 6.As the value of t increases from π/2 to 2π, the particle moves clockwise three-fourths of the way around the ellipse to (x,y) = (2, 5).
Hence, the initial point is (3+sin(π/2), 5+6cos(π/2)) = (3+1, 5) = (4, 5) and the final point is (3+sin(2π), 5+6cos(2π)) = (3, 5)Therefore, the ordered pairs are:(4, 5) → (4, -1) → (2, -1) → (2, 5) The motion of the particle is shown below.
Therefore, the ordered pairs are:(4, 5) → (4, -1) → (2, -1) → (2, 5).
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The point P(1/4,20) 20 lies on the curve y = 5/x. If Q is the point( x,5/x) find the slope of the secant
line PQ for the following values of x.
If x = 0.35, the slope of PQ is:
and if x 0.26, the slope of PQ is:
and if x = 0.15, the slope of PQ is:
and if x = 0.24, the slope of PQ is:
Based on the above results, guess the slope of the tangent line to the curve at P(0.25, 20).
The slope of the tangent line to the curve at P(0.25, 20).
The point P(1/4,20) lies on the curve y = 5/x.
If Q is the point (x,5/x) and we have to find the slope of the secant line PQ for various values of x.
We have the following values of x:If x = 0.35, the slope of PQ is:
(5/0.35 - 20) / (0.35 - 0.25)
= (14.29 - 20) / 0.1= -57.1
If x = 0.26, the slope of PQ is: (5/0.26 - 20) / (0.26 - 0.25)= (19.23 - 20) / 0.01= -77.7
If x = 0.15, the slope of PQ is: (5/0.15 - 20) / (0.15 - 0.25)= (33.33 - 20) / -0.1= -133.3
If x = 0.24, the slope of PQ is: (5/0.24 - 20) / (0.24 - 0.25)= (20.83 - 20) / -0.01= -83.3
Based on the above results, guess the slope of the tangent line to the curve at P(0.25, 20).
We can observe that as we take x closer to 0.25, the slope of the secant line PQ is decreasing, and from the above calculations, we can guess that the slope of the tangent line to the curve at P(0.25, 20) is approximately -80.
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Perform the summation below using the following set of data: \( 2,3,3,4,6 \). \[ \sum\left(4 x^{2}\right)+5 \]
The summation of the given expression
[tex]\(\sum\left(4x^{2}\right)+5\)[/tex] using the set of data {2,3,3,4,6} is 301.
Given set of data is {2,3,3,4,6}.
The summation can be found as follows,
[tex]\[ \sum\left(4x^{2}\right)+5\][/tex]
Let us substitute each element of the given data set in the expression of the summation one by one.
We get
[tex]\[\begin{aligned}&=\left(4\cdot 2^{2}\right)+\left(4\cdot 3^{2}\right)+\left(4\cdot 3^{2}\right)+\left(4\cdot 4^{2}\right)+\left(4\cdot 6^{2}\right)+5 \\&=4\left(4+9+9+16+36\right)+5 \\&=4\cdot 74+5 \\&=296+5 \\&=301\end{aligned}\][/tex]
Hence, the summation of the given expression
[tex]\[ \sum\left(4x^{2}\right)+5\][/tex] using the set of data {2,3,3,4,6} is 301.
This can be concluded as the final answer of the problem.
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The formula to calculate the summation of data is as follows:
\[\sum_{i=1}^{n} x_{i}\]
Whereas the expression given in the question is,
\[\sum\left(4 x[tex]x^{2}[/tex]{2}\right)+5\]
Therefore, as the set of data given is {2, 3, 3, 4, 6}, we can use the formula mentioned above and substitute the values of x.
In this case, n = 5, as we have 5 elements in the data set.
Now, \[\sum_{i=1}^{5} x_{i}\]
Substituting the values,\[\sum_{i=1}^{5} x_{i}=2+3+3+4+6=18\]
Therefore, the value of summation of the given data set is 18.
Now, let's solve for the given expression:
\[\sum\left(4 x^{2}\right)+5\]
On substituting the values from the set of data,
\[\begin{aligned}\sum\left(4 x^{2}\right)+5&=4(2)^{2}+4(3)^{2}+4(3)^{2}+4(4)^{2}+4(6)^{2}+5 \\&=16+36+36+64+144+5 \\&=301\end{aligned}\]
Therefore, the value of the given expression after performing the summation of the given set of data is 301.
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Find the vertex of f(t)=5t2 + 2 t−3 by completing the square. Show your work step by step. Find the value of b such that f(x)=−2x2 +bx+4 has vertex on the line y=x. A farmer finds there is a linear relationship between the number of bean stalks, n, she plants and the yield, y, each plant produces. When she plants 30 stalks, each plant yields 25oz of beans. When she plants 33 stalks, each plant produces 20oz of beans. Find a linear relationship in the form of y=mn+b that gives the yield when n stalks are planted. A farmer wishes to enclose two pens with fencing, as shown below. If the farmer has 700 ft of fencing to work with, what dimensions will maximize the area enclosed? (Do not use) derivatives. Solve this problem using Precalculus knowledge.)
Vertex of f(t)=5t2 + 2 t−3 by completing the square f(t)
= 5t² + 2t - 3To find the vertex of f(t) using the completing the square method, follow these steps:Step 1: Divide the coefficient of the first-degree term by 2 and square the result.
5/2
= 2.5
⇒ (2.5)²
= 6.25
f(t) = 5(t² + 2/5t + 6.25/5 - 6.25/5)
f(x) = -2x² + bx + 4 = x-2x² + bx + 4 - x
= 0-bx - 2x² + x + 4
= 0
The perimeter of the pens is given by:2x + y = 700 - y x
= (700 - y)/2The area of the pens is given by:x y
= (700y - y²)/4To find the maximum area enclosed, we differentiate the above expression and equate it to zero. y
= [tex](700 ± √(700² - 4(4)(-1)(-y²)))/8[/tex]Substituting x
= (700 - y)/2 and y = 25 in the above equation, we obtain:x
= (700 - 25)/2 = 337.5The dimensions that maximize the area enclosed are 337.5 ft and 25 ft.
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) convert the polar equation r = 4 cos θ to rectangular form. identify the curve.
The curve described by this equation is a limacon (a type of cardioid) with a loop.
To convert the polar equation r = 4 cos θ to rectangular form, we use the following relations:
x = r cos θ and y = r sin θ
Substituting r = 4 cos θ, we get:
x = 4 cos θ cos θ = 4 cos^2 θ
y = 4 cos θ sin θ = 2 sin 2θ
Therefore, the rectangular form of the polar equation r = 4 cos θ is x = 4 cos^2 θ and y = 2 sin 2θ.
We can simplify x = 4 cos^2 θ by using the identity cos 2θ = 2 cos^2 θ - 1. Substituting this into the equation above, we get:
x = 2(2 cos^2 θ - 1) = 8 cos^2 θ - 2
So the rectangular form of the polar equation r = 4 cos θ is x = 8 cos^2 θ - 2 and y = 2 sin 2θ.
The curve described by this equation is a limacon (a type of cardioid) with a loop.
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Find the solution of the following initial value problem. y ′′ + y = δ(t − 2π) cost; y(0) = 0, y′ (0) = 1
The solution of the given initial value problem is y(t) = sin(t) + H(t-2π)cos(t-2π), where H(t) is the Heaviside step function.
To solve the initial value problem, we start by finding the complementary solution, which satisfies the homogeneous differential equation y'' + y = 0. The complementary solution is given by y_c(t) = A sin(t) + B cos(t), where A and B are constants to be determined.
Next, we find the particular solution for the given non-homogeneous term δ(t-2π)cos(t). Since the forcing term is a Dirac delta function at t = 2π, we can write the particular solution as y_p(t) = K(t-2π)cos(t-2π), where K is a constant to be determined.
Applying the initial conditions y(0) = 0 and y'(0) = 1, we can solve for the constants A, B, and K. Plugging in these initial conditions into the general solution, we find A = 0, B = 1, and K = 1.
Therefore, the solution of the initial value problem is y(t) = sin(t) + H(t-2π)cos(t-2π), where H(t) is the Heaviside step function.
The initial value problem with the given conditions is solved by finding the complementary solution and the particular solution. The solution is y(t) = sin(t) + H(t-2π)cos(t-2π), where H(t) is the Heaviside step function. This solution satisfies the given initial conditions y(0) = 0 and y'(0) = 1.
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8) Sketch \( f(x)=x^{2}+2 \) on \( [0,3] \). Subdivide the interval into 6 sub-intervals of equal length and compute the upper and lower Riemann sums of \( \mathrm{f} \) over this partition.
The upper Riemann sum of \( f(x) = x^2 + 2 \) over the interval \([0,3]\) with 6 equal sub-intervals is 38, and the lower Riemann sum is 14.
To compute the Riemann sums, we need to divide the interval \([0,3]\) into 6 sub-intervals of equal length. The length of each sub-interval will be \(\Delta x = \frac{3-0}{6} = \frac{1}{2}\).
For the upper Riemann sum, we evaluate the function at the right endpoints of each sub-interval and multiply it by the width of the sub-interval. The right endpoints for the 6 sub-intervals are \(x = \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}, 3\). Evaluating \(f(x) = x^2 + 2\) at these points, we get \(f(\frac{1}{2}) = \frac{9}{4} + 2 = \frac{17}{4}\), \(f(1) = 3\), \(f(\frac{3}{2}) = \frac{13}{4} + 2 = \frac{21}{4}\), \(f(2) = 6\), \(f(\frac{5}{2}) = \frac{29}{4} + 2 = \frac{37}{4}\), and \(f(3) = 11\). Multiplying each of these function values by the width of the sub-interval \(\frac{1}{2}\) and summing them up, we get the upper Riemann sum of 38.
For the lower Riemann sum, we evaluate the function at the left endpoints of each sub-interval and multiply it by the width of the sub-interval. The left endpoints for the 6 sub-intervals are \(x = 0, \frac{1}{2}, 1, \frac{3}{2}, 2, \frac{5}{2}\). Evaluating \(f(x) = x^2 + 2\) at these points, we get \(f(0) = 2\), \(f(\frac{1}{2}) = \frac{1}{4} + 2 = \frac{9}{4}\), \(f(1) = 3\), \(f(\frac{3}{2}) = \frac{9}{4} + 2 = \frac{17}{4}\), \(f(2) = 6\), and \(f(\frac{5}{2}) = \frac{25}{4} + 2 = \frac{33}{4}\). Multiplying each of these function values by the width of the sub-interval \(\frac{1}{2}\) and summing them up, we get the lower Riemann sum of 14.
The upper Riemann sum of 38 and the lower Riemann sum of 14 provide upper and lower estimates for the area under the curve of \(f(x) = x^2 + 2\) over the interval \([0,3]\) using a partition with 6 equal sub-intervals.
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A fisherman can clean a fish according to a Poisson (get it!?!?) process with a rate of 1 every 20 seconds, once the catch is landed he gets to work immediately. Use the Central Limit Theorem to estimate the probability that the 6th fish is finished being cleaned between 103 and 139 seconds after he starts.
The probability that the sixth fish is finished being cleaned between 103 and 139 seconds after he starts is approximately 0.7498.
As per the given question,
A fisherman can clean a fish according to a Poisson process with a rate of 1 every 20 seconds.
Let X be the time taken by the fisherman to clean a single fish.
The rate of the Poisson process is λ = 1/20 per second.
So, the mean of X is μ = E(X) = 20 seconds and the variance of X is σ[tex]-^{2}[/tex] = Var(X) = 400 [tex]s^{2}[/tex].
Now, the time taken by the fisherman to clean n fishes, [tex]Y_{n}[/tex], can be modeled by a gamma distribution with parameters n and λ, i.e., [tex]Y_{n}[/tex] ~ Gamma(n,λ).
The mean of [tex]Y_{n}[/tex] is
μ' = E([tex]Y_{n}[/tex]) = n/λ seconds
and the variance of [tex]Y_{n}[/tex] is
σ'[tex]-^{2}[/tex] = Var([tex]Y_{n}[/tex]) = n/λ[tex]-^{2}[/tex] [tex]s^{2}[/tex].
By the central limit theorem, [tex]Y_{n}[/tex] can be approximated by a normal distribution with mean μ' and variance σ'^2/n, i.e.,[tex]Y_{n}[/tex] ~ N(μ',σ'[tex]-^{2}[/tex]/n).
So, the time taken by the fisherman to clean 6 fishes, [tex]Y_{6}[/tex], can be approximated by a normal distribution with mean
μ' = E([tex]Y_{6}[/tex]) = 6/λ = 120 seconds
and
variance σ'[tex]-^{2}[/tex]/n = Var([tex]Y_{6}[/tex])/6 = 100/6 [tex]s^{2}[/tex].
The probability that the sixth fish is finished being cleaned between 103 and 139 seconds after he starts can be calculated as follows:
Z[tex]^{1}[/tex] = (103 - μ')/[tex]\sqrt{}[/tex](σ'[tex]-^{2}[/tex]/6) = -1.15
Z[tex]^{2}[/tex] = (139 - μ')/[tex]\sqrt{}[/tex](σ'[tex]-^{2}[/tex]/6) = 1.15
Using the standard normal distribution table or calculator, the probability can be found as:
P(-1.15 < Z < 1.15) = P(Z < 1.15) - P(Z < -1.15) = 0.8749 - 0.1251 = 0.7498
Therefore, the probability that the sixth fish is finished being cleaned between 103 and 139 seconds after he starts is approximately 0.7498.
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find sin 2a and tan 2a if cos a = -0.6, pi/2 < a < pi
The values of the required trigonometric identities are:
sin 2α = -0.96
tan 2α = 3.43
How to solve trigonometric identities?In trigonometric identities we know that in the second quadrant, that sin is positive while cosine and tangent are negative.
cos α = -0.6
This can be written as: -3/5
Using Pythagoras theorem, the other side is 4 and as such:
sin α = 4/5 = 0.8
We know that:
sin 2α = 2sin α cos α
Thus:
sin 2α = 2 * 0.8 * -0.6
= -0.96
tan 2α = sin 2α/(cos²α - sin²α)
Thus:
tan 2α = -0.96/((-0.6)² - 0.8²)
tan 2α = 3.43
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Brigid is picking strawberries at the Pick-Your-Own Farm. Her goal is to pick 5 bushels of strawberries. She has already picked 1
1
2
bushels, and she picks at a rate of
5
8
bushel per hour. The scenario is represented as
5
8
h + 1
1
2
= 5, where h is the number of hours she picks. How many more hours will it take Brigid to fill 5 bushels of strawberries?
2 and StartFraction 3 Over 16 EndFraction hours
2 and StartFraction 3 Over 16 EndFraction hours
5 and three-fifths hours
10 and two-fifths hours
Answer:
We can start by isolating the variable "h".
5 8 h + 1 1 2 = 5
Subtracting 11/2 from both sides:
5 8 h = 5 - 1 1 2
Simplifying:
5 8 h = 8 1 2
Dividing both sides by 5/8:
h = 8 1 2 ÷ 5 8
Converting the mixed number to an improper fraction:
h = (8 x 8 + 1) ÷ 5 8
h = 65/8
Now, we can convert this fraction to a mixed number:
h = 8 1/8
Brigid has already picked for 8 1/8 hours, so the amount of time needed to pick the remaining strawberries is:
5 - (1 1/2 + 5/8 x 8) = 5 - (3 5/8) = 1 3/8
Therefore, Brigid still needs to pick for 1 3/8 hours to fill 5 bushels of strawberries. The answer is 1 and 3/8 hours or 2 and 3/16 hours (if simplified).
Step-by-step explanation:
The barometric pressure y (in inches of mercury) at an altitude of x miles above sea level decreases at a rate proportional to the current pressure according to the model
dx
dy
=−0.2y where y=29.92 inches when x=0. (Round your answers to two decimal places.) (a) Find the barometric pressure at 9372 feet. in (b) Find the barometric pressure at 19,510 feet. in
a. The barometric pressure at 9372 feet is approximately 25.89 inches of mercury.
b. The barometric pressure at 19,510 feet is approximately 5.82 inches of mercury.
To solve the differential equation dx/dy = -0.2y,
separate variables and integrate.
(a) Finding the barometric pressure at 9372 feet,
First, convert 9372 feet to miles.
Since 1 mile is equal to 5280 feet, we have,
9372 feet = 9372/5280 miles ≈ 1.774 miles (rounded to three decimal places).
x = 1.774,
The initial condition y = 29.92 when x = 0.
Now, let's solve the differential equation,
dx/dy = -0.2y
Separating variables,
dy/y = -0.2dx
Integrating both sides,
∫(1/y)dy = -0.2∫dx
⇒ln|y| = -0.2x + C
Applying the initial condition (x = 0, y = 29.92),
⇒ln|29.92| = -0.2(0) + C
⇒ln|29.92| = C
Therefore, the equation becomes,
ln|y| = -0.2x + ln|29.92|
To find the barometric pressure at 9372 feet (x = 1.774), substitute the value into the equation,
ln|y| = -0.2(1.774) + ln|29.92|
Now, solve for y by exponentiating both sides of the equation,
|y| = [tex]e^{(-0.2(1.774)[/tex]+ ln|29.92|)
Since the barometric pressure cannot be negative,
Take the positive value,
y = [tex]e^{(-0.2(1.774)[/tex] + ln|29.92|)
Calculating the value numerically
y ≈ 25.89 inches (rounded to two decimal places)
(b) Finding the barometric pressure at 19,510 feet,
Convert 19,510 feet to miles,
19,510 feet
= 19,510/5280 miles
≈ 3.69 miles (rounded to two decimal places).
x = 3.69,
The initial condition y = 29.92 when x = 0.
ln|y| = -0.2x + ln|29.92|
Substituting x = 3.69 into the equation,
ln|y| = -0.2(3.69) + ln|29.92|
Solving for y,
y = [tex]e^{(-0.2(3.69)[/tex]+ ln|29.92|)
Calculating the value numerically,
y ≈ 5.82 inches (rounded to two decimal places)
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represent the plane curve by a vector-valued function. y = x 5
The vector-valued function that represents the plane curve y = [tex]x^5[/tex] can be written as r(t) = (t, [tex]t^5[/tex]), where t is a parameter that represents the parameterization of the curve.
We have,
In vector calculus, a vector-valued function is a function that takes a parameter and outputs a vector.
In this case, we want to represent the plane curve y = [tex]x^5[/tex] as a vector-valued function.
The curve y = [tex]x^5[/tex] is a polynomial equation relating the y-coordinate to the x-coordinate.
To represent this curve as a vector-valued function, we can assign the parameter t to the x-coordinate and express the y-coordinate in terms
of t.
Using the vector notation, we can write the vector-valued function as r(t) = (x(t), y(t)), where x(t) and y(t) represent the x and y coordinates of the curve as functions of the parameter t.
In our case, we can assign t to x, so we have x(t) = t.
Then, we can express y(t) in terms of t as y(t) = (t^5). Combining these, we get r(t) = (t, [tex]t^5[/tex]).
Thus,
The vector-valued function that represents the plane curve y = [tex]x^5[/tex] is r(t) = (t, [tex]t^5[/tex]), where t is the parameter representing the parameterization of the curve.
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find the horizontal asymptote. give your answer in the form y=b, or enter ∅ if there is none. f(x)=15−x
The horizontal asymptote of the function f(x) = 15 - x is y = -1.
To find the horizontal asymptote of the function f(x) = 15 - x, we need to determine the behavior of the function as x approaches positive or negative infinity.
For this linear function, the degree of the numerator and denominator is the same (both are of degree 1). In such cases, the horizontal asymptote can be determined by looking at the coefficient of the highest power term.
In the function f(x) = 15 - x, the coefficient of the highest power term (x¹) is -1. Therefore, the horizontal asymptote is determined by the ratio of the leading coefficients, which is -1/1 or simply -1.
The horizontal asymptote is y = -1.
Therefore, the horizontal asymptote of the function f(x) = 15 - x is y = -1.
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Find a basis for the left-nullspace for
\( A=\left[\begin{array}{ccc}1 & 3 & 2 \\ 0 & -1 & 2 \\ 3 & 8 & 8 \\ 2 & 4 & 8\end{array}\right] \)
A basis for the left-nullspace of matrix A is given by the vector [3, 1, 1, -1].
To find a basis for the left-nullspace of matrix A, we need to find the vectors x that satisfy the equation xA = 0, where 0 is the zero vector. In other words, we are looking for vectors x that, when multiplied from the left by matrix A, result in the zero vector.
We can find the left-nullspace by finding the nullspace of the transpose of A (A^T). This is because the nullspace of A^T is equivalent to the left-nullspace of A.
Let's calculate the left-nullspace basis using the following steps:
1. Compute the transpose of matrix A:
A^T = [1 0 3 2
3 -1 8 4
2 2 8 8]
2. Reduce the augmented matrix [A^T | 0] to row-echelon form:
[1 0 3 2 | 0;
0 1 -1 -2 | 0;
0 0 0 0 | 0]
3. Express the remaining variables (leading variables) in terms of the free variables (non-leading variables) to obtain the left-nullspace basis.
In this case, there is one free variable, which we can assign as t. Therefore, the left-nullspace basis can be represented as follows:
x = t * [3; 1; 1; -1]
Hence, a basis for the left-nullspace of matrix A is given by the vector [3, 1, 1, -1].
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Construct a 3x3 matrix A and a nonzero vector b such that b is in Col. A, but b is not the same as any one of the columns of A.
The 3x3 matrix A =
[tex]\[\begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \\\end{bmatrix}\][/tex] and the nonzero vector b = [tex]\[\begin{bmatrix}1 \\1 \\1 \\\end{bmatrix}\][/tex] satisfy the conditions.
Vector b lies in the column space of A since it can be expressed as a linear combination of the columns of A. However, b is not the same as any of the columns of A as it has different values for each entry.
To construct a 3x3 matrix A and a nonzero vector b such that b is in the column space (Col. A), but b is not the same as any one of the columns of A, we can create a matrix with linearly dependent columns.
Let's consider the following matrix A and vector b:
A = [tex]\[\begin{bmatrix}1 & 2 & 1 \\2 & 4 & 2 \\3 & 6 & 3 \\\end{bmatrix}\][/tex]
b = [tex]\[\begin{bmatrix}3 \\6 \\9 \\\end{bmatrix}\][/tex]
In this case, we can observe that vector b is in the column space of matrix A since it is a linear combination of the columns of A. However, b is not equal to any one of the columns of A.
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Given f(x)=4x3−x−7, what is the value of x in the
interval [1, 1.5] for which the function takes the value 1?
The value of x in the interval [1, 1.5] for which f(x) = 1 is approximately 1.408.
To find the value of x in the interval [1, 1.5] for which the function f(x) = 4x³ - x - 7 takes the value 1, we can use a numerical method like the bisection method or Newton's method.
Let's use the bisection method to solve this problem.
The bisection method works by repeatedly bisecting the interval and narrowing it down until we find a solution within a desired tolerance.
In this case, we want to find a value of x such that f(x) = 1. We'll start with the interval [a, b] = [1, 1.5] and iteratively narrow it down.
Let's perform the iterations using the bisection method:
Iteration 1:
a = 1, b = 1.5
c = (a + b) / 2 = (1 + 1.5) / 2 = 1.25
f(c) = 4(1.25)³ - 1.25 - 7 ≈ -1.422
Since f(c) is negative, the solution lies in the right half of the interval [1.25, 1.5].
Iteration 2:
a = 1.25, b = 1.5
c = (a + b) / 2 = (1.25 + 1.5) / 2 = 1.375
f(c) = 4(1.375)³ - 1.375 - 7 ≈ -0.287
Since f(c) is still negative, the solution still lies in the right half of the interval [1.375, 1.5].
Iteration 3:
a = 1.375, b = 1.5
c = (a + b) / 2 = (1.375 + 1.5) / 2 = 1.4375
f(c) = 4(1.4375)³ - 1.4375 - 7 ≈ 0.360
Since f(c) is now positive, the solution lies in the left half of the interval [1.375, 1.4375].
Iteration 4:
a = 1.375, b = 1.4375
c = (a + b) / 2 = (1.375 + 1.4375) / 2 = 1.40625
f(c) = 4(1.40625)³ - 1.40625 - 7 ≈ -0.467
Since f(c) is still negative, the solution still lies in the right half of the interval [1.40625, 1.4375].
Continuing this process, we can narrow down the interval further until we reach the desired tolerance.
After performing several more iterations, we find that the value of x in the interval [1, 1.5] for which f(x) = 1 is approximately 1.408.
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34 $4000 is invested into an account paying interest at 8%, compounded annually and an extra $200
is invested atter each 12 months. Thus:
Amount in account at end of 1 yr
= $4000 x 1.08 + $200
+ T,
Amount in account at end of 2 yrs
= ($4000 × 1.08 + $200) x 1.08 + $200 + Tz
Express In+ in terms of T, and determine (nearest cent) the amount in the account at the end
of ten years, after the $200 for that year has been added.
The amount in the account at the end of ten years, after the $200 for that year has been added is given by Amount in account after 10 years= $6539.81 (nearest cent).
The given expression for the amount in the account at the end of 1 year, Amount in account at end of
1 yr= $4000 × 1.08 + $200 + T,
and that for the amount in the account at the end of 2 years, Amount in account at end of
2 yrs= ($4000 × 1.08 + $200) × 1.08 + $200 + T2
can be generalized as follows: Amount in account at end of n years, where n is a positive integer,
we have
In= $4000 × 1.0810−1 + $200 {13.9332 + T[1 − (1.08)9/0.08]}.
Now, substituting T=0, we obtain, Amount in account after
10 years= $4000 × 1.0810−1 + $200 {13.9332 + 0[1 − (1.08)9/0.08]}.
Thus, the amount in the account at the end of ten years, after the $200 for that year has been added is given by Amount in account after 10 years= $6539.81 (nearest cent).
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. Find a number c such that the parabola y = 2x ^2 + 8x + 2c − 6
has its vertex on the x-axis
The number c = 9 such that the vertex of the parabola y = 2x^2 + 8x + 2c - 6 lies on the x-axis.
To find the number c such that the vertex of the parabola y = 2x^2 + 8x + 2c - 6 lies on the x-axis, we need to set the y-coordinate of the vertex equal to zero. The y-coordinate of the vertex of a parabola in the form y = ax^2 + bx + c is given by -b/2a.
In this case, we have a = 2 and b = 8. Setting -b/2a = 0, we get:
-(8) / (2*2) = 0
Simplifying, we have:
-8 / 4 = 0
This implies that the parabola's vertex lies at x = 0. Now we substitute x = 0 into the equation of the parabola:
y = 2(0)^2 + 8(0) + 2c - 6
y = 0 + 0 + 2c - 6
y = 2c - 6
Since we want the vertex to lie on the x-axis, the y-coordinate should be equal to zero. Therefore, we set y = 0:
0 = 2c - 6
Adding 6 to both sides and dividing by 2, we find:
2c = 6
c = 6/2
c = 3
The number c that satisfies the given condition is c = 3.
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Find the absolute maximum and absolute minimum values of f on the given interval. (Round all answers to two decimal places.)
f(t) = t + cot(t/2)
[pi/4, (7 pi)/4]
The absolute maximum value is approximately 5.71, and the absolute minimum value is approximately 1.57 for the function f(t) = t + cot(t/2) on the interval [π/4, (7π)/4].
To find the absolute maximum and absolute minimum values of the function f(t) = t + cot(t/2) on the interval [π/4, (7π)/4], we can follow these steps:
Find the critical points of the function by taking the derivative of f(t) and setting it equal to zero.
Evaluate the function at the critical points and the endpoints of the interval.
Compare the values obtained to determine the absolute maximum and absolute minimum.
Let's start by finding the critical points:
Step 1:
Taking the derivative of f(t) with respect to t:
f'(t) = 1 - (1/2)[tex]cosec^2(t/2)[/tex]
Setting f'(t) equal to zero:
1 - (1/2)[tex]cosec^2(t/2)[/tex] = 0
Solving for t:
[tex]cosec^2(t/2)[/tex] = 2
[tex]sin^2(t/2)[/tex] = 1/2
sin(t/2) = ±√(1/2)
t/2 = π/4 or t/2 = (3π)/4
t = π/2 or t = (3π)/2
So, the critical points are t = π/2 and t = (3π)/2.
Step 2:
Now, we evaluate the function f(t) at the critical points and the endpoints of the interval:
f(π/4) = (π/4) + cot((π/4)/2) = (π/4) + 1 = 1 + π/4 ≈ 1.79
f((7π)/4) = ((7π)/4) + cot(((7π)/4)/2) = ((7π)/4) - 1 = 7π/4 - 1 ≈ 5.71
f(π/2) = (π/2) + cot((π/2)/2) = (π/2) + 0 = π/2 ≈ 1.57
f((3π)/2) = ((3π)/2) + cot(((3π)/2)/2) = ((3π)/2) + 0 = 3π/2 ≈ 4.71
Step 3:
Comparing the values, we can determine the absolute maximum and absolute minimum:
Absolute maximum: The maximum value is f((7π)/4) ≈ 5.71, which occurs at t = (7π)/4.
Absolute minimum: The minimum value is f(π/2) = 1.57, which occurs at t = π/2.
Therefore, the absolute maximum value is approximately 5.71, and the absolute minimum value is approximately 1.57 for the function f(t) = t + cot(t/2) on the interval [π/4, (7π)/4].
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Find f. \[ f(x)-B+6 x+24 x^{2}, \quad f(0)=2, \quad f(1)-13 \]
The value of function f is f(x) = -2 + 6 x + 24 x².
Given information is as follows:
f(x)-B+6 x+24 x^{2}
From question, we know that,
f(0) = 2
Let's find f(0) by putting x = 0 in the given equation as follows:
\[f(0) - B + 6(0) + 24(0)^2 = 2\]
or,
\[f(0) - B = 2\]
or,
\[B = f(0) - 2\]
So, B = 0 - 2
= -2
Now, we know that
f(1)-13
Let's find f(1) by putting x = 1 in the given equation as follows:
\[f(1) - (-2) + 6(1) + 24(1)^2 = 13\]
or,
\[f(1) = 13 - 2 - 6 - 24 = -19\]
So, \[f(x) = -2 + 6 x + 24 x^2\]
Conclusion: Therefore, the value of f is f(x) = -2 + 6 x + 24 x².
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set up an integral for the length of the curve. b. graph the curve to see what it looks like. c. use a grapher's or computer's integral evaluator to find the curve's length numerically.
Graphing the curve helps visualize its shape, and using a computer's integral evaluator allows for numerical computation of the curve's length.a. The integral for the length of the curve is:
∫[pi/5 to 4pi/5] sqrt(1 + (2cos(y))^2) dy
b. The correct graph would be a sinusoidal curve with increasing amplitude as y ranges from pi/5 to 4pi/5.
c. Using a graphing calculator or online integral evaluator, the numerical value for the length of the curve is approximately 2.426 units.
To calculate the length of a curve, we start by setting up an integral that represents the arc length. Given a function y = f(x) over a certain interval [a, b], the arc length integral is defined as:
L = ∫[a, b] √(1 + (f'(x))^2) dx,
where f'(x) is the derivative of the function f(x) with respect to x. This formula accounts for the infinitesimal lengths along the curve.
Next, graphing the curve allows us to visually examine its shape and understand its behavior. Plotting the function y = f(x) on a coordinate plane provides a visualization of the curve, giving insights into its curvature, steepness, and any notable features.
Finally, to find the curve's length numerically, we can use a grapher's or computer's integral evaluator. This tool allows us to input the arc length integral formula and the specific function we want to evaluate. The integral evaluator then computes the definite integral to find the numerical value of the curve's length.
By utilizing these steps - setting up the integral, graphing the curve, and using an integral evaluator - we can accurately determine the length of a curve numerically.
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Approximate the value of f"(2.156) if f(x) = 2tan(x) + cos(2x). h = 0.003 -18.22610955 8.396938164 O 8.424277328 -18.51527191 Approximate the value of f"(7.585) if f(x) = ecos(3x). h = 0.003 O 5.248377254 5.323581886 O 5.399963657 5.477487216
The approximate value of f"(2.156) if f(x) = 2tan(x) + cos(2x) with h = 0.003 is -18.22610955.
The approximate value of f"(7.585) if f(x) = ecos(3x) with h = 0.003 is 5.323581886.
To approximate the value of the second derivative of a function using finite differences, we can use the central difference formula. The central difference formula for the second derivative is given by:
f"(x) ≈ (f(x + h) - 2f(x) + f(x - h))/h²
where h is a small step size.
For the function f(x) = 2tan(x) + cos(2x), we need to approximate f"(2.156). Using h = 0.003, we can apply the central difference formula:
f"(2.156) ≈ (f(2.156 + 0.003) - 2f(2.156) + f(2.156 - 0.003))/(0.003)²
Substituting the function values into the formula and calculating the result, we find that f"(2.156) is approximately -18.22610955.
For the function f(x) = ecos(3x), we need to approximate f"(7.585). Again, using h = 0.003, we can apply the central difference formula:
f"(7.585) ≈ (f(7.585 + 0.003) - 2f(7.585) + f(7.585 - 0.003))/(0.003)²
Substituting the function values into the formula and calculating the result, we find that f"(7.585) is approximately 5.323581886.
Therefore, the approximate value of f"(2.156) if f(x) = 2tan(x) + cos(2x) with h = 0.003 is -18.22610955, and the approximate value of f"(7.585) if f(x) = ecos(3x) with h = 0.003 is 5.323581886.
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Locate all absolute maxima and minima, if any, by inspection. Then check your answers using calculus: \[ f(x, y)=(x-7)^{2}+(y+8)^{2} \]
f has an ______ at _________
The function f has an absolute minimum at (7, -8).
To locate the absolute maxima and minima of the function f(x, y) = [tex](x-7)^2[/tex] + [tex](y+8)^2[/tex], we can first inspect the function visually and then confirm our findings using calculus.
By inspecting the function, we notice that it represents a paraboloid centered at the point (7, -8) with the vertex at that point. Since the function is a sum of squares, both terms [tex](x-7)^2[/tex] and [tex](y+8)^2[/tex] are always non-negative. Therefore, the function f(x, y) is also non-negative for all values of x and y.
Since the function is non-negative, there are no absolute minima. However, there is an absolute maximum at the vertex (7, -8) since the function attains its minimum value of zero at that point.
To confirm our findings using calculus, we can find the critical points of the function. Taking partial derivatives with respect to x and y, we have:
∂f/∂x = 2(x-7)
∂f/∂y = 2(y+8)
Setting these derivatives equal to zero to find the critical points, we have:
2(x-7) = 0 => x = 7
2(y+8) = 0 => y = -8
Thus, the only critical point is (7, -8), which matches our inspection.
To determine the nature of this point, we can use the second partial derivative test. Taking the second partial derivatives:
∂²f/∂x² = 2
∂²f/∂y² = 2
Since both second partial derivatives are positive, the point (7, -8) is a local minimum.
Therefore, the function f(x, y) = [tex](x-7)^2[/tex] + [tex](y+8)^2[/tex] has an absolute minimum at the point (7, -8), confirming our initial inspection.
In summary, the function has an absolute minimum at (7, -8) and no absolute maxima.
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(1 point) use stokes' theorem to evaluate ∬s(∇×f)⋅ds where f(x,y,z)=−3yzi 3xzj 12(x2 y2)zk and s is the part of the paraboloid z=x2 y2 that lies inside the cylinder x2 y2=1, oriented upward.
The value of the integral using Stokes' theorem is -4/3π.
To use Stokes' theorem, we need to find the curl of f:
∇×f =
| i j k |
| ∂/∂x ∂/∂y ∂/∂z |
| -3yz 3xz 12(x^2-y^2) |
= (6xy-3x)i + (-3z)j + (-2x^2-2y^2)k
Now, we need to parametrize the surface s. We can use cylindrical coordinates by letting x = rcosθ, y = rsinθ, and z = r^2, where 0 ≤ θ ≤ 2π and 0 ≤ r ≤ 1.
The unit normal vector to the surface is given by:
n = (∂z/∂r × ∂z/∂θ)/|∂z/∂r × ∂z/∂θ|
= (-2r cosθ, -2r sinθ, 1)/sqrt(4r^2+1)
The integral we want to evaluate is then:
∬s(∇×f)⋅ds = ∫∫curl(f)⋅n dS
= ∫∫[(6xy-3x)(-2r cosθ) + (-3r^2)(-2r sinθ) + (-2r^2)(-2r^2)]/sqrt(4r^2+1) dA
= ∫∫(12r^3sinθ - 6r^2cosθ - 8r^5)/sqrt(4r^2+1) dA
We can evaluate this integral using polar coordinates by letting x = rcosθ, y = rsinθ, and dA = r dr dθ:
∫∫(12r^3sinθ - 6r^2cosθ - 8r^5)/sqrt(4r^2+1) dA
= ∫₀²π ∫₀¹ (12r^4sinθ - 6r^3cosθ - 8r^5)/sqrt(4r^2+1) dr dθ
= ∫₀²π [(-2/3)(4r^2+1)^(-3/2) (12r^4sinθ - 6r^3cosθ - 8r^5)]|₀¹ dθ
= (-2/3π)[3sinθ + 3cosθ]|₀²π
= -4/3π
Therefore, the value of the integral using Stokes' theorem is -4/3π.
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Identify the quadric surface by name. Find and describe the xy-, xz-, and yz-traces, when they exist.
16x^2-16y^2-z=0
A. Hyperbolic paraboloid; xz-trace: 16x^2-z=0 (parabola); yz-trace: 16y² + z = 0 (parabola)
B. Elliptic paraboloid; xz-trace: 16x²-z=0 (parabola); yz-trace: 16y²+z = 0 (parabola)
C. Elliptic paraboloid; xz-trace: 16x² -z = 0 (hyperbola); yz-trace: 16y² + z = 0 (hyperbola)
D. Elliptic cone; xz-trace: 16x2 -z=0 (hyperbola); yz-trace: 16y²+z=0 (hyperbola)
A. Hyperbolic paraboloid; xz-trace: [tex]16x^2-z=0[/tex] (parabola); yz-trace: [tex]16y^2 + z = 0[/tex] (parabola)
To identify the quadric surface, we can analyze the equation:
[tex]16x^2 - 16y^2 - z = 0[/tex]
This equation represents a hyperbolic paraboloid.
To find the xy-, xz-, and yz-traces, we set one variable to zero and solve for the other two variables:
1. xy-trace (z = 0):
[tex]16x^2 - 16y^2 = 0[/tex]
Simplifying, we get:
[tex]x^2 - y^2 = 0[/tex]
This is a difference of squares, which factors as:
(x - y)(x + y) = 0
So the xy-trace consists of two lines: x = y and x = -y.
2. xz-trace (y = 0):
[tex]16x^2 - z = 0[/tex]
Solving for z, we have:
[tex]z = 16x^2[/tex]
This represents a parabola opening upwards in the xz-plane.
3. yz-trace (x = 0):
[tex]-16y^2 - z = 0[/tex]
Solving for z, we get:
[tex]z = -16y^2[/tex]
This represents a parabola opening downwards in the yz-plane.
Therefore, the xy-trace consists of two lines (x = y and x = -y), the xz-trace is a parabola opening upwards, and the yz-trace is a parabola opening downwards.
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consider polar functions r = 1 cos(θ) and r = 1 − cos(θ), 0 ≤ θ ≤ 2π. find the area of the shaded region.
The area of the shaded region bounded by the polar functions r = 1 cos(θ) and r = 1 − cos(θ), 0 ≤ θ ≤ 2π is 5/2 + 3/4π - 3√2/4.
To find the area of the shaded region bounded by the polar functions r = 1 cos(θ) and r = 1 − cos(θ), 0 ≤ θ ≤ 2π, we need to perform the following steps:
Graph the two polar functions Find the points of intersection between the two functions
Set up the integral for the area of the shaded region
Integrate the function over the given interval to find the area of the shaded region.
Graph of the two polar functions r = 1 cos(θ) and r = 1 − cos(θ), 0 ≤ θ ≤ 2π is shown below:
The two polar functions intersect at θ = π/2 and θ = 3π/2 (as shown in the graph).To find the area of the shaded region,
we need to set up the integral as follows:∫(from 0 to π/2) [1/2 (1−cos θ)² - 1/2 (1 cos θ)²]dθ + ∫(from π/2 to 3π/2) [1/2 (1−cos θ)² - 1/2 (1 cos θ)²]dθ + ∫(from 3π/2 to 2π) [1/2 (1−cos θ)² - 1/2 (1 cos θ)²]dθ.
Simplifying the integral, we get:∫(from 0 to π/2) [1/2 - cos θ + 1/2 cos² θ]dθ + ∫(from π/2 to 3π/2) [1/2 - cos θ + 1/2 cos² θ]dθ + ∫(from 3π/2 to 2π) [1/2 - cos θ + 1/2 cos² θ]dθ.
On integrating, we get: 1/4 [ 7 sin (θ) - sin (2θ) + 2θ ] from 0 to π/2 + 1/4 [ 7 sin (θ) - sin (2θ) + 2θ ] from π/2 to 3π/2 + 1/4 [ 7 sin (θ) - sin (2θ) + 2θ ] from 3π/2 to 2π.
Substituting the limits in the above expression, we get:Area of the shaded region = 5/2 + 3/4π - 3√2/4So, this is the main answer for the given problem which is computed as per the steps above. Answer more than 100 words are provided above.
The conclusion is that the area of the shaded region bounded by the polar functions r = 1 cos(θ) and r = 1 − cos(θ), 0 ≤ θ ≤ 2π is 5/2 + 3/4π - 3√2/4.
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Find the indicated quantity. 6) Find the moment of inertia with respect to its axis of the solid generated by revolving the region bounded by y=4x−x
2
and the x-axis about the y-axis, where k is the density.
The moment of inertia with respect to its axis of the solid generated by revolving the region is 227.62k.
To find the moment of inertia of a solid generated by revolving a region about an axis, we can use the formula:
I = ∫(r² × dm)
I is the moment of inertia
r is the distance from the axis of rotation to the element of mass
dm is the mass element
We are revolving the region bounded by y = 4x - x² and the x-axis about the y-axis.
To find the limits of integration, we need to determine the x-values where the curve intersects the x-axis. Setting y = 0, we can solve for x:
0 = 4x - x²
x² - 4x = 0
x(x - 4) = 0
Therefore, our limits of integration will be from x = 0 to x = 4.
To express the equations in terms of x, we solve y = 4x - x² for x:
4x - x² = y
x² - 4x + y = 0
Now, let's find the mass element dm. Since k is the density, we can express dm as:
dm = k × dV
where dV is the infinitesimal volume element. In this case, dV will be equal to the area element, which can be expressed as:
dV = A × dx
where A is the area enclosed by the curve at a given x-value.
To find the area A at a given x-value, we integrate y with respect to x:
A = ∫(y) dx
Plugging in y = 4x - x², we have:
A = ∫(4x - x²) dx
= [2x² - (1/3)x³] evaluated from x = 0 to x = 4
= 10.67
dm = k × A × dx
I = ∫(r² × dm)
= ∫(r² × k × A × dx)
The distance from the y-axis to the element of mass is simply x, so r = x.
I = ∫(x² × k × A) dx
= k × A × ∫(x²) dx
= k × A × [(1/3)x³] evaluated from x = 0 to x = 4
= k × A × (64/3)
Now we substitute the value of A we found earlier:
I = k ×10.67 × (64/3)
= 227.62k
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determine whether the series is convergent or divergent by expressing sn as a telescoping sum (as in thisexample). Σn->3 -[infinity] 2/(n^2 − 1) n
Since sn = 1/2 - 1/(n + 2), as n → ∞, sn → 1/2. Therefore, the given series converges to 1/2.
The given series is Σn=3∞ 2/(n² - 1)n. We need to express sn as a telescoping sum. Let's start by finding a general formula for the nth term of the series, tn.
tn = 2/(n² - 1)n = 2/[(n - 1)(n + 1)]n.
The given expression can be written as:
Σn=3∞ 2/(n² - 1)n
= Σn=3∞ [1/ (n - 1) - 1/(n + 1)]
Multiplying numerator and denominator of the first term by (n + 1) and the second term by (n - 1), we get
Σn=3∞ [1/ (n - 1) - 1/(n + 1)]
= [1/2 - 1/4] + [1/3 - 1/5] + [1/4 - 1/6] + .......+ [1/n - 1/(n + 2)] + .......
Now, let's find a formula for the nth partial sum, sn.
s1 = [1/2 - 1/4]
s2 = [1/2 - 1/4] + [1/3 - 1/5]
s3 = [1/2 - 1/4] + [1/3 - 1/5] + [1/4 - 1/6]......
s2 = [1/2 - 1/4] + [1/3 - 1/5]
s3 = [1/2 - 1/4] + [1/3 - 1/5] + [1/4 - 1/6]....+ [1/n - 1/(n + 2)]
s3 - s2 = [1/4 - 1/6] + [1/5 - 1/7] + [1/6 - 1/8].....- [1/3 - 1/5]
s4 - s3 = [1/5 - 1/7] + [1/6 - 1/8] + [1/7 - 1/9].....- [1/4 - 1/6]
s4 - s1 = [1/3 - 1/5] + [1/4 - 1/6] + [1/5 - 1/7].....- [1/n - 1/(n + 2)]
It can be observed that, on simplifying sn, the terms get cancelled, leaving only the first and last terms. Hence, sn can be written as:sn = [1/2 - 1/(n + 2)]
Therefore, the given series is a telescoping series, which means that each term after a certain point cancels out with a previous term, leaving only the first and last terms.
Since sn = 1/2 - 1/(n + 2), as n → ∞, sn → 1/2. Therefore, the given series converges to 1/2.
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Environmental physics (do them all please) A homogeneous water-sand suspension inside a cubic barrel (side length L = 1 m). If the sand particles have a settling velocity Vrs = 10 mm/s and the water is stagnant inside the barrel: (a) What kind of particle motion (laminar or turbulent) while settling in the barrel? (b) How long does it take for the water to be cleared out of the sand particles? (c) If the water is stirred in the barrel, how long does it take for the sand concentration to drop to 10% of its initial concentration? Question 7: A filter is installed in a pipe with an air flow rate 30 m³/hour to collect an aerosol sample. After one 24 hours, the total mass collected on the filter was 2.88 mg. The collected particles have diameter 10 µm and density 1.7 g/cm³. 1 (a) What is the average mass concentration during this sampling? (b) What is the average number concentration during this sampling?
The particle motion in the water-sand suspension is laminar due to the relatively low settling velocity of the sand particles. The time it takes for the water to be cleared out of the sand particles is approximately 100 seconds.
(a) The particle motion in the water-sand suspension while settling in the barrel can be considered laminar. This is because the settling velocity of the sand particles is relatively low (10 mm/s), indicating a slow and smooth motion through the water. In laminar flow, particles move in ordered layers without significant mixing or turbulence. The low settling velocity suggests that the forces of gravity and viscosity dominate the particle motion, resulting in a laminar flow regime.
(b) To determine how long it takes for the water to be cleared out of the sand particles, we can use the settling velocity and the height of the barrel. Since the settling velocity is given as 10 mm/s, it would take 1 meter / 10 mm/s = 100 seconds (or 1 minute and 40 seconds) for the sand particles to settle to the bottom of the barrel.
(c) If the water is stirred in the barrel, the rate at which the sand concentration drops will depend on the intensity of the stirring and the mixing dynamics. Generally, stirring enhances the mixing and accelerates the decrease in sand concentration. The time required for the sand concentration to drop to 10% of its initial concentration will vary based on the specific conditions and parameters of the stirring process, such as the stirring speed, duration, and the initial sand concentration.
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What is the value of n in the equation 1/2 (n-4)-3=3-(2n+3)
Answer:
n = 2
Step-by-step explanation:
(n -4)/2 -3 = 3 -(2n +3)
(n -4)/2 -3 = 3 -2n -3
(n -4)/2 -3 = -2n
(n -4)/2 = -2n +3
n -4 = 2(-2n +3)
n -4 = -4n +6
5n = 10
n = 2
Consider the equation r=
6+12sinθ
8
. Find the eccentricity, identify the conic, and give an equation of the directrix. (5 pta.) Find a polar equation for an ellipse with eccentricity
4
1
and directrix y=2 whose focus is at the origin
The eccentricity of the equation is √7/4, representing an ellipse, with the directrix equation y = -2, and the polar equation r = (√7/2)(1 - (4/7)cosθ) for the given ellipse.
The eccentricity of the equation r = (6 + 12sinθ)/8 is √(1 - (b²/a²)) = √(1 - (3²/4²)) = √(1 - 9/16) = √(7/16) = √7/4.
The equation represents an ellipse since the eccentricity (e) is between 0 and 1.
The equation of the directrix is y = -c = -2.
The polar equation for an ellipse with eccentricity 4/√7 and directrix y = 2 and the focus at the origin is r = a(1 - e²)/(1 + ecosθ), where a = distance from the origin to the focus = 2/e = 2/(4/√7) = √7/2.
Therefore, the polar equation is r = (√7/2)(1 - (4/7)cosθ).
Learn more about the polar equation at
https://brainly.com/question/28976035
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