Question

Suppose that the length l, width w, and perimeter P of a rectangle are differentiable functions of t. Suppose also that the length is always three times and twenty-two more than the width. Write an equation that relates dPdt to dwdt. Enter your answers using exact values.

Answer 1

Answer:

$\dfrac{dP}{dt}=8\dfrac{dw}{dt}$

Step-by-step explanation:

The perimeter, P, is given by

$P(t) = 2(l(t)+w(t))$

Since $l(t) = 3w(t) + 22$,

$P(t) = 2(3w(t) + 22+w(t)) = 2(4w(t)+22) = 8w(t)+44$

Differentiating P with respect to w,

$\dfrac{dP}{dw} = 8$

But

$\dfrac{dP}{dw} = \dfrac{dP}{dt}\times \dfrac{dt}{dw} = \dfrac{dP}{dt}/\dfrac{dw}{dt}$

$\dfrac{dP}{dt} = \dfrac{dP}{dw} \times \dfrac{dw}{dt} = 8\dfrac{dw}{dt}$

Answer 2

The equation that relates (dP/dt) to (dp/dw) is;

dp/dt = 8(dw/dt)

We are told that;

Length of rectangle is l

width of rectangle is w

Perimeter of rectangle is P

Formula for perimeter of rectangle is;

P = 2l + 2w

We are told that the length is always three times and twenty-two more than the width. Thus;

l = 3w + 22.

∴ P = 2(3w + 22) + 2w

P = 6w + 44 + 2w

P = 8w + 44

Differentiating with respect to w gives;

dp/dw = 8

Now, we are told that they are all differentiable functions of t and so;

P(t) = 8w(t) + 22

We want to relate (dP/dt) with (dp/dw) and we have;

dP/dt = (dp/dw) × (dw/dt)

dp/dt = 8(dw/dt)

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