Question

The quantities $P, Q$ and $R$ are functions of time and are related by the equation $R=P Q$. Assume that $P$ is increasing instantaneously at the rate of $8 \%$ per year and that $Q$ is decreasing instantaneously at the rate of $2 \%$ per year. That is, $\frac{P^{\prime}}{P}=0.08$ and $\frac{Q^{\prime}}{Q}=-0.02$. Determine the percentage rate of change for $R$.

   The quantities $P, Q$ and $R$ are functions of time and are related by the equation $R=P Q$. Assume that $P$ is increasing instantaneously at the rate of $8 \%$ per year and that $Q$ is decreasing instantaneously at the rate of $2 \%$ per year. That is, $\frac{P^{\prime}}{P}=0.08$ and $\frac{Q^{\prime}}{Q}=-0.02$. Determine the percentage rate of change for $R$.
 
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CLP-1 Differential Calculus 1
CLP-1 Differential Calculus 1
Joel Feldman, Andrew… 1st Edition
Chapter 3, Problem 3 ↓
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The quantities $P, Q$ and $R$ are functions of time and are related by the equation $R=P Q$. Assume that $P$ is increasing instantaneously at the rate of $8 \%$ per year and that $Q$ is decreasing instantaneously at the rate of $2 \%$ per year. That is, $\frac{P^{\prime}}{P}=0.08$ and $\frac{Q^{\prime}}{Q}=-0.02$. Determine the percentage rate of change for $R$.
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Transcript

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00:01 Now the key of this problem is knowing the product rule, where they have r of t is defined as p of t times q of t.
00:11 So i'm going to do this in green because that's what we have going on here.
00:16 And i know a lot of people do this differently, but i'm going to try and say consistent that r prime of t you leave alone and then you take the derivative of q and plus take the derivative of p times you leave q alone.
00:33 I know some people do that a little bit differently, but that's okay.
00:37 So r prime of we're doing 100 would be equal to p of 100.
00:45 So i'm just going to the numbers.
00:47 It says p of 100 is 4 .65...
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