1. A rectangle has a constant area of 200 square meters and its length L is increasing at the rate of 4 meters per second.
a. Find the width W at the instant the width is decreasing at the rate of .4 meters per second.
b. At what rate is the diagonal D of the rectangle changing at the instant when the width W is 10 meters?
To solve this problem, we need to use the formulas for the area of a rectangle and the Pythagorean theorem.
a. To find the width W at the instant when it is decreasing at the rate of 0.4 meters per second, we will differentiate the area formula and solve for the width.
The area of a rectangle is given by the equation A = L * W, where A is the area, L is the length, and W is the width. Since the area is constant at 200 square meters, we have:
200 = L * W
Now, we need to differentiate both sides of this equation with respect to time (t) to find the rates of change of the variables. Since L is increasing at a rate of 4 meters per second, we have dL/dt = 4. Similarly, we are given that dW/dt = -0.4 (since the width is decreasing at a rate of 0.4 meters per second).
Differentiating the equation with respect to time:
d(200) / dt = d(L * W) / dt
0 = dL/dt * W + L * dW/dt
Substituting the given values:
0 = 4 * W + 200 * (-0.4)
Now, solve for W:
0 = 4W - 80
4W = 80
W = 20
Therefore, at the instant when the width is decreasing at a rate of 0.4 meters per second, the width W of the rectangle is 20 meters.
b. To find the rate of change of the diagonal D of the rectangle when the width W is 10 meters, we can use the Pythagorean theorem.
The Pythagorean theorem states that the square of the hypotenuse (D) of a right triangle is equal to the sum of the squares of the other two sides (L and W). Therefore:
D^2 = L^2 + W^2
Differentiating both sides of this equation with respect to time (t):
2D * dD/dt = 2L * dL/dt + 2W * dW/dt
We can now substitute the known values into the equation:
D = √(L^2 + W^2) = √(L^2 + 10^2)
dL/dt = 4 (given)
dW/dt = ? (what we need to find)
L = ? (what we need to find)
W = 10 (given)
Since the area of the rectangle is constant at 200 square meters, we have:
L * W = 200
Therefore, L = 200 / 10 = 20
Now, substitute these values into the equation and solve for dD/dt:
2D * dD/dt = 2L * dL/dt + 2W * dW/dt
2D * dD/dt = 2(20) * 4 + 2(10) * dW/dt
Simplifying the equation:
dD/dt = (40 + 20 * dW/dt) / D
Now, substitute D = √(20^2 + 10^2) = √500 into the equation:
dD/dt = (40 + 20 * dW/dt) / √500
Finally, substitute the given value dW/dt = -0.4 and calculate the rate:
dD/dt = (40 + 20 * (-0.4)) / √500
dD/dt = (40 - 8) / √500
dD/dt = 32 / √500
Simplify the square root:
dD/dt = 32 / (10 * √5)
dD/dt = 16 / √5
Therefore, at the instant when the width W is 10 meters, the rate of change of the diagonal D of the rectangle is approximately 16 / √5 meters per second.