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?
a. Ah, the shrinking rectangle. Let's solve this clownishly!
We know that the area of a rectangle is given by A = L * W, and it's constant at 200 square meters. So, we have 200 = L * W.
Now, we need to find the rate at which the width is decreasing, which is given as -0.4 meters per second.
To find the width W at that instant, we can take the derivative of the equation 200 = L * W with respect to time (t) and set it equal to -0.4:
0 = L * dW/dt + W * dL/dt
Since dL/dt = 4 meters per second, we have:
0 = L * dW/dt + 4W
Now, we know that L is increasing at a rate of 4 meters per second, so we can substitute that in:
0 = (4t) * dW/dt + 4W
Simplifying, we get:
dW/dt = -W / t
Now we plug in the rate at which the width is decreasing (-0.4 meters per second):
-0.4 = -W / t
Solving for W, we have:
W = 0.4t
So, at the instant the width is decreasing at the rate of 0.4 meters per second, the width is given by 0.4t.
b. Ah, the changing diagonal! Here we go!
Let's first find an expression for the diagonal D of the rectangle in terms of length L and width W.
Using the Pythagorean theorem, we have:
D^2 = L^2 + W^2
Differentiating implicitly with respect to time (t), we get:
2D * dD/dt = 2L * dL/dt + 2W * dW/dt
Given that the width W is 10 meters, we can substitute it in:
2D * dD/dt = 2L * 4 + 20 * dW/dt
We already know that dW/dt is given as -0.4 meters per second.
Now, we're asked to find the rate of change of the diagonal at the instant when the width is 10 meters. So, we need to find dD/dt at that instant.
Plugging in the values, we have:
2D * dD/dt = 8L - 8
Substituting the area equation (200 = L * W) in terms of L, we have:
2D * dD/dt = 8(200 / W) - 8
Since W = 10 meters, we get:
2D * dD/dt = 8(200 / 10) - 8
2D * dD/dt = 80 - 8
2D * dD/dt = 72
Dividing both sides by 2D, we get:
dD/dt = 36 / D
So, at the instant when the width is 10 meters, the rate of change of the diagonal is given by 36 divided by the diagonal itself.
Step 1: Find the relationship between the length L and the width W of the rectangle.
The area of a rectangle is given by the formula A = L * W, where A is the area, L is the length, and W is the width.
Step 2: Use the given information to set up an equation to solve for the width W at the instant the width is decreasing at the rate of 0.4 meters per second.
Given that the area is constant at 200 square meters, we have A = 200. We are also given that the length is increasing at a rate of 4 meters per second, so dL/dt = 4. And we are asked to find dW/dt when dW/dt = -0.4 (the width is decreasing at 0.4 meters per second).
Step 3: Use the equation to solve for the width W.
We can rearrange the equation A = L * W to solve for W: W = A / L.
Step 4: Substitute the given values into the equation to find W.
Since A = 200 and dL/dt = 4, we have W = 200 / (L + 4).
Step 5: Substitute the given values into the equation W = 200 / (L + 4) and solve for W when dW/dt = -0.4.
Substituting dL/dt = 4 and dW/dt = -0.4, we have -0.4 = 200 / (L + 4). Solve this equation for L.
Step 6: Solve the equation -0.4 = 200 / (L + 4) for L.
To solve this equation, we can multiply both sides by L + 4: -0.4(L + 4) = 200.
Step 7: Simplify the equation and solve for L.
Expanding the left side of the equation, we get -0.4L - 1.6 = 200.
Then, add 1.6 to both sides of the equation to isolate -0.4L: -0.4L = 201.6.
Finally, divide both sides of the equation by -0.4 to solve for L: L = 201.6 / -0.4.
Step 8: Calculate the width W at the instant the width is decreasing at the rate of 0.4 meters per second.
Substitute the value of L into the equation W = 200 / (L + 4):
W = 200 / (201.6 / -0.4 + 4).
Step 9: Simplify the equation to find the width W.
Simplifying the equation, we get:
W = 200 / (504 - 8) = 200 / 496.
Therefore, the width of the rectangle at the instant when the width is decreasing at the rate of 0.4 meters per second is W ≈ 0.402 meters.
Step 10: Find the relationship between the length L, the width W, and the diagonal D of the rectangle.
The diagonal of a rectangle can be found using the Pythagorean theorem. The relationship between L, W, and D is given by the equation D = sqrt(L^2 + W^2).
Step 11: Use the given information to set up an equation to find the rate at which the diagonal D is changing.
At the instant when the width W is 10 meters, we want to find the rate at which the diagonal D is changing. We are given dW/dt = 0 (since the width is not changing), and we are asked to find dD/dt.
Step 12: Use the equation D = sqrt(L^2 + W^2) to find the derivative dD/dt.
Differentiate both sides of the equation D = sqrt(L^2 + W^2) with respect to time: dD/dt = (1/2)(L^2 + W^2)^(-1/2)(2LdL/dt + 2WdW/dt).
Step 13: Substitute the given values into the equation dD/dt = (1/2)(L^2 + W^2)^(-1/2)(2LdL/dt + 2WdW/dt).
Substituting the given values L = 201.6 and W = 10, and dL/dt = 4 and dW/dt = 0, we have: dD/dt = (1/2)(201.6^2 + 10^2)^(-1/2)(2(201.6)(4) + 2(10)(0)).
Step 14: Simplify the equation to find the rate at which the diagonal D is changing.
Simplifying the equation, we get: dD/dt = (1/2)(40416 + 100)^(-1/2)(806.4) = (1/2)(40516)^(-1/2)(806.4).
Therefore, the rate at which the diagonal D of the rectangle is changing at the instant when the width W is 10 meters is approximately dD/dt ≈ 0.010 meters per second.
To solve these problems, we will use the formulas for the area, width, and diagonal of a rectangle.
1. For a rectangle, the area is given by the formula:
A = length * width
Given that the area is constant at 200 square meters, we have:
200 = L * W
To find the width W at the instant when the width is decreasing at 0.4 meters per second, we need to find the derivative of the equation with respect to time and solve for the width when the derivative is -0.4 m/s.
Differentiating both sides of the equation with respect to time t, we get:
d(200) / dt = d(L * W) / dt
Since the area is constant, the derivative of 200 with respect to time is 0. The derivative of L * W with respect to time can be found using the product rule of differentiation:
0 = (dL/dt) * W + L * (dW/dt)
We know that dL/dt = 4 m/s (given), and we need to find dW/dt when it is -0.4 m/s. Substituting the values into the equation, we have:
0 = 4 * W - 0.4 * L
Since we have a constant area, we can express the length in terms of the width:
L = 200 / W
Substituting this into the equation, we get:
0 = 4 * W - 0.4 * (200 / W)
Simplifying the equation, we have:
4W = 0.4 * (200 / W)
Multiplying both sides by W, we get:
4W^2 = 0.4 * 200
Solving for W, we have:
W^2 = (0.4 * 200) / 4
W^2 = 8
Taking the square root of both sides, we get:
W = sqrt(8)
W ≈ 2.83 meters
Therefore, at the instant when the width is decreasing at a rate of 0.4 meters per second, the width of the rectangle is approximately 2.83 meters.
2. To find the rate at which the diagonal D of the rectangle is changing at the instant when the width W is 10 meters, we can use the Pythagorean theorem, which relates the length, width, and diagonal of a rectangle:
D^2 = L^2 + W^2
Taking the derivative of both sides of the equation with respect to time t, we get:
2D * (dD/dt) = 2L * (dL/dt) + 2W * (dW/dt)
Since we want to find dD/dt when W = 10, we need to find the values of D, L, dL/dt, and dW/dt at that instant. We know that W = 10, and we can express L in terms of W using the area equation as before:
200 = L * W
L = 200 / W
L = 200 / 10
L = 20 meters
Similarly, we have:
dL/dt = 4 m/s (given)
dW/dt = ? (unknown)
Substituting these values into the derivative equation, we get:
2D * (dD/dt) = 2 * 20 * 4 + 2 * 10 * (dW/dt)
2D * (dD/dt) = 160 + 20 * (dW/dt)
At this point, we don't have enough information to solve the equation because we don't know the value of D or dW/dt. We need either of these values to find the rate of change of the diagonal.
lw=200
l dw/dt+ W dl/dt=0
given dl/dt=4
a) find w then dw/dt=-.4
-.4*L+W*4=0
but remember LW=200 or L=200/W
-.4*200/W+4W=0
solve for W
b. diagonal D.
D= sqrt(L^2+W^2
dD/dt= 1/2 *1/sqrt( ) * (2L dl/dt+ 2W dw/dt)