3.9 Related Rates

Question

Suppose y = sqrt2x+1 where x and y are functions of t.

(a)

If

dx

dt

= 9,

find

dy

dt

when

x = 4.

dy

dt

=

3

(b)

If

dy

dt

= 4,

find

dx

dt

when

x = 24.

dx

dt

=

Answer 1

The problem statement you gave is a bit garbled. I will give you a rundown of how to solve the problem. Basically, you want to compare rates of change when it comes the the function y and the function x. ie. you are looking to compare dy/dt and dx/dt. Because x is a function of t, when we take the derivative of the function y with respect to t we need to use the chain rule: dy/dt = dy/dx • dx/dt

For example suppose we have the function y = x², and we want to find dy/dt when x = 2 and dx/dt = 1/2.

We apply the chain rule to y, so dy/dt = dy/dx • dx/dt.

dy/dx = 2x, and since we don't know the inner workings of the function x(t), we simply have dx/dt as a place holder for itself.

The chain rule dy/dt = dy/dx • dx/dt becomes

dy/dt = 2x dx/dt. Since we are looking for dy/dt, when x = 2 and dx/dt = 1/2, we simply plug those values in AFTER we have applied the chain rule to find the derivative dy/dt.

We get dy/dt = 2 (2) (1/2).

So dy/dt = 2

I am unsure of which of the following your function is

1. y = √(2) x + 1
2. y = √(2x) + 1
3. y = √(2x+1)

Part (a) also looks like it is already answered from my interpretation of this question.

I will solve part b for you assuming that y = √(2x+1)

dy/dt = 4 and we want to solve for dx/dt when x = 24.

First we find the derivative of y with respect to t (dy/dt). Again, we need to use the chain rule here.

dy/dt = dy/dx • dx/dt

dy/dt = 1/√(2x+1) • dx/dt

Since we know that dy/dt = 4 and x = 24, we can simply plug those values in.

4 = 1/√(25) • dx/dt

we simply solve by multiplying 5 on both sides and we get

20 = dx/dt

Answer 2

a. y = √(2x+1) When x = 4, y = √9 = 3.

dy/dt = (1/2)(2x+1)^-1/2(2)(dx/dt) = (1 / √(2x+1))(dx/dt)

When x = 4 and dx/dt = 9, we have dy/dt = (1 / √9)(9) = 3.

Do part b in a similar manner.