Let A be the area if the region in the first quadrant under the graph of
y = cos(x) and above the line y = k for
0 < k < 1.
a. Determine A in terms of k.
b. Determine the value of A when k = 1/2
c. If the line y = k is moving upward at a rate of 1/pi units per minute, at what rate is the area A changing when k = 1/2?
a. To find the area A in terms of k, we need to find the x-values where the graph of y = cos(x) intersects the line y = k.
The graph of y = cos(x) is periodic with a period of 2π and symmetric about the y-axis. It intersects the x-axis at x = (2n + 1/2)π, where n is an integer.
To find the points of intersection between y = cos(x) and y = k, we set the equations equal to each other:
cos(x) = k
Taking the inverse cosine (arccos) of both sides gives:
x = arccos(k)
Since we only want the region in the first quadrant, we need to find the positive value of x. The area under the graph of y = cos(x) and above the line y = k is bounded by x = 0 and x = arccos(k).
Therefore, the area A in terms of k is:
A = ∫[0, arccos(k)] cos(x) dx
b. To find the value of A when k = 1/2, we substitute k = 1/2 into the integral:
A = ∫[0, arccos(1/2)] cos(x) dx
We can evaluate this integral using trigonometric identities:
A = sin(arccos(1/2))
A = sin(π/3)
A = √3/2
c. To find the rate at which the area A is changing with respect to time, we need to differentiate the integral with respect to k and then multiply by the rate at which k is changing.
Let A' be the derivative of A with respect to k, and dk/dt be the rate at which k is changing. Then the rate of change of A with respect to time is given by:
dA/dt = A' * dk/dt
To find A', we differentiate the integral:
A' = d/dk ∫[0, arccos(k)] cos(x) dx
Using the Second Fundamental Theorem of Calculus, we have:
A' = cos(arccos(k)) * d(arccos(k))/dk
A' = √(1 - k^2) * (-1/√(1 - k^2))
A' = -1
Since dk/dt = 1/π, we can substitute the values into the rate of change equation:
dA/dt = -1 * (1/π)
dA/dt = -1/π
To find the area A in terms of k, we need to take the integral of the function y = cos(x) in the interval where it lies above the line y = k.
a. The integral of the function y = cos(x) in the interval where it lies above y = k can be written as:
A = ∫[0, π/2] (cos(x) - k) dx
To evaluate this integral, we integrate the function cos(x) - k with respect to x from 0 to π/2:
A = [sin(x) - kx] evaluated from 0 to π/2
A = (sin(π/2) - k(π/2)) - (sin(0) - k(0))
A = 1 - k(π/2)
So, A in terms of k is A = 1 - k(π/2).
b. To determine the value of A when k = 1/2, we substitute k = 1/2 into the equation we found in part a:
A = 1 - (1/2)(π/2)
A = 1 - π/4
Therefore, when k = 1/2, A = 1 - π/4.
c. To find the rate at which the area A is changing with respect to time when k = 1/2, we need to take the derivative of A with respect to k and then multiply it by the rate at which k is changing with respect to time.
First, let's find dA/dk:
dA/dk = d/dk (1 - k(π/2))
dA/dk = -π/2
Now, let's find the rate at which k is changing with respect to time, dk/dt. We are given that k is moving upward at a rate of 1/π units per minute. Therefore:
dk/dt = 1/π
Finally, to find the rate at which A is changing when k = 1/2, we multiply dA/dk by dk/dt:
dA/dt = (dA/dk) * (dk/dt)
dA/dt = (-π/2) * (1/π)
dA/dt = -1/2
Therefore, when k = 1/2, the rate at which the area A is changing is -1/2 units per minute.