# Differentials where the variable undergoes a percentage increase. Where am I wrong?

Let $R = \frac{k}{r^4}$, where $k$ is some constant. Find the change in $R$ as $r$ is increased by 10%. $R$ is the resistance of blood flow, $r$ is the radius of a vein.

This problem seems easy enough to do with some simple algebra, but I'm lead to believe (based on context) that this problem should be attempted using differentials. However, when I attempt the problem using differentials I get a completely different answer.

Method 1: $R_0 = \frac{k}{r_0^4}$ (our original resistance). $R=\frac{k}{r^4}$ (our new resistance). Let $r=1.1r_0$. Then $\frac{R}{R_0} = \frac{k/r^4}{k/r_0^4}=\frac{r_0}{r}=\frac{1}{1.1^4} \approx .683$. This tells us that an increase by 10 percent in our radius results in a $31.7$% decrease in $R$. This is what I would wager is the correct answer.

Method 2: $\frac{dR}{dr} = \frac{-4k}{r^5}$ Thus, $dR=\frac{-4k}{r^5}*dr$. Now, usually I'm given some fact like, $r$ goes from 3 to 2.98 or something like that, where $\Delta r$ is explicit. In this case though, our $\Delta r = .1r$. $$\Delta R \approx \frac{-4k}{r^5}*\Delta r=-.4\frac{k}{r^4}=-.4R$$ Here, we see that a 10% increase in the radius should give rise to a 40% decrease in $R$, contradicting our past solution.

Question: Where am I going wrong? I feel like my algebraic method should be exactly correct, but this contradicts my differential method. In the second method, I can see that as $r$ gets larger and larger, our approximation should get worse and worse. Does this have something to do with my contradictory answers?

Both of your methods are correct. The first one gives you the exact decrease in R, while the second is just a first order approximation which becomes better when your change in $r$ is smaller.

• Yes, as long as a derivative exists, or at least one-sided derivative. For example $f(x)=|x|$ is not differentiable at $x=0$, but the one sided derivatives exist and they are -1 and 1. Commented Jul 27, 2015 at 20:45

I suppose that you are astonished because your two results have a different sign. Well, this is because your Method 1, is not correct.

The: change in $R$ as $r$ is increased by $10$% is: $$\Delta R = R-R_0=\dfrac{k}{(1.1 r_0)^4}-\dfrac{k}{ r_0^4}=k\left( \dfrac{1-(1.1)^4}{(1.1)^4r_0^4}\right)$$ and this change relative to $R_0$ is: $$\dfrac{\Delta R}{R_0}=\dfrac{1}{(1.1)^4}-1 \approx-0.32 < 0$$

Your Method 2 is correct but, as noted in the answer of Svetoslav, it is a linear approximation so, for a variation of $0.1$ in the independent variable it gives a result that is different from the exact value because the function is not linear.

• Why would you say my method 1 is incorrect? It gives exactly the same answer as you have given. Namely, they both show a 32% decrease in $R$. Moreover, the two results don't give different signs. They both show that $R$ decreases as $r$ increases. What astonishes me is the degree to which they are different from one another. Commented Jul 27, 2015 at 20:29
• Yes, this was what I was going to say- the first method of @PaddlingGhost is correct, because his sign is in the text - he wrote 31.7 % decrease. Commented Jul 27, 2015 at 20:34
• Sorry! I did not notice the world "decrease" . So my answer is a confirm of the answer of Svetoslav. Commented Jul 27, 2015 at 20:46