# Finding the derivative and rate of change

• ver_mathstats
In summary, at time t = 2, the demand for a good will be -60000 p^(-3) greater than the demand at time t = 1.
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.

ver_mathstats said:
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}##.

tnich said:
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).

WWGD said:
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
ver_mathstats said:
I still got -20.93115678 as my answer.
That is the correct numerical answer. Your equation in post #3 is actually correct.

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

WWGD
ver_mathstats said:
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.

## 1. What is a derivative and how is it calculated?

A derivative is a mathematical concept that represents the instantaneous rate of change of a function at a specific point. It is calculated by finding the slope of the tangent line to the function at that point.

## 2. What is the significance of finding the derivative?

The derivative is important in many areas of science and mathematics because it helps us understand the behavior of a function and how it changes over time or space. It is also used to solve optimization problems and make predictions about the future behavior of a system.

## 3. How is the derivative related to the rate of change?

The derivative and the rate of change are closely related, as the derivative represents the instantaneous rate of change of a function at a specific point. In other words, it tells us how much the function is changing at that exact moment.

## 4. What are some real-life applications of finding derivatives and rates of change?

Finding derivatives and rates of change is used in many fields, such as physics, engineering, economics, and biology. Some examples include determining the velocity of an object in motion, finding the maximum profit for a business, and analyzing the growth rate of a population.

## 5. How can I use calculus to find the derivative and rate of change?

Calculus provides us with the tools and techniques to find derivatives and rates of change. By using the rules of differentiation, such as the power rule and chain rule, we can find the derivative of almost any function. Additionally, we can use the derivative to find the slope of a tangent line and calculate the rate of change at a specific point.

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