# Finding the derivative and rate of change

• ver_mathstats

#### ver_mathstats

Homework Statement
When printer paper is sold for p dollars per packet, consumers will buy D(p)=30000/p^2 units per month. It is estimated that t months from now the price of paper will be p(t)=0.3t^(1/2)+6.3 dollars per unit. At what rate will the monthly demand for the paper be changing with respect to time 2 months from now?
Relevant Equations
D(p)=30000/p^2
p(t)=0.3t^(1/2)+6.3
t=2
I tried solving this question a few ways and this one logically made the most sense however I got it wrong and I am unsure of why.

I first plugged in t=2 into p(t).

p(2)=0.3(2)1/2+6.3 to obtain 6.724264069.

I then found the derivative of D(p) which is D'(p)=-60000/p3.

I plugged in 6.724264069 into D'(p). D'(6.724264069)= -197.34.

I am confused at how to proceed and where I went wrong.

Thank you.

I then found the derivative of D(p) which is D'(p)=-60000/p3.
You are trying to find the rate of change of D(p) with respect to time, so starting with what you did here, you need to use the chain rule to find the derivative of ##D(p)## with respect to time ##t##. The result will have a ##\dot p## in it. Then you can differentiate ##p(t)## with respect to ##t## and substitute the result into your expression for ##\frac {dD(p)}{dt}##.

You are trying to find the rate of change of D(p) with respect to time, so starting with what you did here, you need to use the chain rule to find the derivative of ##D(p)## with respect to time ##t##. The result will have a ##\dot p## in it. Then you can differentiate ##p(t)## with respect to ##t## and substitute the result into your expression for ##\frac {dD(p)}{dt}##.
Thank you for the help. So, I struggled to differentiate it so I am still unsure if I did it correct and ended up with D'p(t)=30000(-2p(t)-3⋅p'(t)).

My answer all together I got was -20.93115678, is this still wrong?

This is the chain rule. You are given D=D(p(t)) then D'(t)=D'(P(t))P'(t).

This is the chain rule. You are given D=D(p(t)) then D'(t)=D'(P(t))P'(t).
I still got -20.93115678 as my answer.

• tnich
I still got -20.93115678 as my answer.
That is the correct numerical answer. Your equation in post #3 is actually correct.

• ver_mathstats
That is the correct numerical answer. Your equation in post #3 is actually correct.
Thank you for the help.

• WWGD
Homework Equations: D(p)=30000/p^2
p(t)=0.3t^(1/2)+6.3
t=2

I tried solving this question a few ways and this one logically made the most sense however I got it wrong and I am unsure of why.

I first plugged in t=2 into p(t).
Substituting for t is NOT the first thing you should do. In fact, it's pretty much the last thing you should do.
You've already found the solution, but I thought I should correct some misgivings on your part.

There are at least two ways to approach this problem.
1. Since demand D is a function of unit price p, and since p is a function of t, you could write D as a function of t, and then differentiate. Once you have D'(t), then and only then should you substitute t = 2, not before.
2. Use the chain rule, as other members have shown in this thread.
##\frac d{dt}\left(D(p)\right) = \frac d{dp}\left(D(p)\right) \cdot \frac {dp}{dt} = D'(p) \cdot p'(t)##
## = -60000p^{-3}(.15t^{-1/2} = -60000(.3t^{1/2} + 6.3)^{-3}(.15t^{-1/2})##
At this point, but not before, you can replace t by 2 to find the rate of change of demand D with respect to time at time t = 2.