A rectangle has width w inches and height h, where the width is twice the height. Both w and h are functions of time,t, measured in seconds. If A represents the area of the rectangle, what is the rate of change of A with respect to t at the instant where the width is 4 inches and the height is increasing at the rate of 2 inches per second?
Ah, the wonderful world of rectangles and rates of change! Let me put on my clown mathematician hat for this one.
We know that the width, w, is twice the height, so we can say that w = 2h.
Now let's express the area, A, of the rectangle in terms of w and h. The area of a rectangle is simply the product of its width and height, so A = w * h.
Since we want to find the rate of change of A with respect to time, we need to take the derivative of A with respect to t.
Now let's substitute the equation for w that we found earlier into the equation for A:
A = (2h) * h
A = 2h^2
To find the rate of change of A with respect to t, we can apply the chain rule of differentiation.
dA/dt = dA/dh * dh/dt
Let's differentiate A = 2h^2 with respect to h first:
dA/dh = 4h
Now let's find dh/dt, which represents the rate at which the height is changing with respect to time. We know that dh/dt is increasing at a rate of 2 inches per second.
Substituting our values,
dA/dt = (4h) * (2)
Now, at the instant where the width is 4 inches, we can substitute w = 4 into our equation w = 2h to find the height, h.
4 = 2h
h = 2 inches
So, at that instant, the height, h, is 2 inches. Substituting this value into our expression for dA/dt:
dA/dt = (4 * 2) * (2)
dA/dt = 16 inches^2/sec
Therefore, at the instant where the width is 4 inches and the height is increasing at the rate of 2 inches per second, the rate of change of the area, A, with respect to time, t, is 16 inches^2/sec.
Hope that helps, and remember, math can be funny too!
To find the rate of change of the area of the rectangle with respect to time, we need to use the chain rule in calculus.
Let's start by expressing the width and height in terms of t.
Given that the width is twice the height, we can write:
w = 2h
The area of the rectangle is given by the formula:
A = w * h
Substituting the expression for w, we have:
A = (2h) * h
A = 2h^2
Now, let's differentiate both sides of the equation with respect to t:
dA/dt = d(2h^2)/dt
Using the chain rule, we have:
dA/dt = 2 * d(h^2)/dt
To find d(h^2)/dt, we need to differentiate h^2 with respect to t.
Using the power rule, we have:
d(h^2)/dt = 2h * dh/dt
Given that the height is increasing at a rate of 2 inches per second (dh/dt = 2), we can substitute this value into the equation:
d(h^2)/dt = 2h * 2
d(h^2)/dt = 4h
Now, substituting this value back into the previous equation for dA/dt, we have:
dA/dt = 2 * 4h
dA/dt = 8h
At the instant where the width is 4 inches, we know that the width is twice the height. So if the width is 4 inches, the height would be 2 inches.
Substituting h = 2 into the equation for dA/dt, we have:
dA/dt = 8 * 2
dA/dt = 16
Therefore, the rate of change of the area with respect to time at the instant where the width is 4 inches and the height is increasing at a rate of 2 inches per second is 16 square inches per second.
To find the rate of change of the area of the rectangle with respect to time, we need to differentiate the area function with respect to time.
Let's start by expressing the width and height in terms of time. We are given that the width, w, is twice the height, h. So, we can write:
w = 2h
Next, let's express the area, A, of the rectangle in terms of the width and height. The area of a rectangle is given by the formula:
A = w * h
Substituting the value of w in terms of h, we get:
A = (2h) * h = 2h^2
Now, we have the area of the rectangle, A, expressed in terms of the height, h.
To find the rate of change of A with respect to time, we need to differentiate A with respect to t. Differentiating 2h^2 with respect to t, we get:
dA/dt = d(2h^2)/dt = 4h * dh/dt
Now, we have the rate of change of A with respect to t in terms of h and dh/dt.
Given that the height is increasing at a rate of 2 inches per second (dh/dt = 2), and we want to find the rate of change of A at the instant where the width is 4 inches (w = 4), we need to find the value of h at that instant.
From the given information that w = 4 and w = 2h, we can solve for h:
4 = 2h
h = 2 inches
Now, we can substitute the values of h and dh/dt in the expression for dA/dt:
dA/dt = 4h * dh/dt
dA/dt = 4 * 2 * 2
dA/dt = 16 square inches per second
Therefore, the rate of change of the area of the rectangle with respect to time at the instant where the width is 4 inches and the height is increasing at the rate of 2 inches per second is 16 square inches per second.
width = twice height
w = 2h or h = w/2
area = wh = 2h^2
A = 2h^2
dA/dt = 4h dh/dt
for the given data ... when w = 4, h = 2
dA/dt = 4(2)(2) inches^2 /sec
= 16 inches^2/sec