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I was just given this problem in class and it seems to me I don't have enough information. $V=16r$ and $V=4r^2$ would both satisfy the given derivative, but $\Delta r=0.1$ would produce different changes in $V$ depending on the function. Did my professor just forget to give a piece of information, am I not understanding the problem, or is perhaps the spirit of the questions more general than my approach. Is it just asking me to solve this equation: $\Delta V = 16 \Delta r$ when $\Delta r = 0.1$? Thanks for reading.

UPDATE

As Toby Mak mentions in the answer below, we can view the problem as $dV=16dr$ where an "infinitesimal nudge" to $r$ translates to 16 times that nudge to $dV$ (when $r=2$). However, when we move from infinitesimals to concrete nudges like $\Delta r=0.1$, I don't see how we could actually know whether or not we have a good approximation. As we don't have the original function, we have no error bounds, so a nudge of $0.1$ (though we generally consider this to be a small nudge) really isn't different from a nudge of 1000000. To illustrate this, consider a function $$ V(r) = \left\{ \begin{array}{ll} 16r & \quad r \leq 2.05 \\ 1000000r & \quad r > 2.05 \end{array} \right. $$ Here we have $V(2)=32$ but $V(2.1)=2100000$. So our change in $V$ was nowhere near the estimate of $1.6$. Though this function was constructed to suit my needs, if $V$ can be chosen from the set of all possible functions with a derivative of $16$ at $r=2$, then most functions won't have a $\Delta V$ that's close to 1.6.

So this brings me back to wondering what the point of this questions is, and does an approximation of $\Delta V$ make sense without having more information?

Sean
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  • It does at $r=2$. In other words $V'(2)=16$. – Sean Feb 08 '20 at 02:22
  • Sorry for the mistake. Something's wrong with your approach but I can't figure out what that is. I'm just about to post an answer to your question. – Toby Mak Feb 08 '20 at 02:25
  • Found it: the issue with your approach is that when $\Delta r$ approaches $0$, $\Delta V$ will be the same for both functions. But $\Delta V$ is a number greater than $0$, so $\Delta V$ will be different from the actual derivative. – Toby Mak Feb 08 '20 at 02:31

1 Answers1

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Is it just asking me to solve this equation: $\Delta V = 16 \Delta r$ when $\Delta r = 0.1$?

Exactly. But it is also important to understand the conceptual reason as to why that is.

First, look at $\frac{\mathrm d V}{\mathrm d r} = 16$. What this tells us is when $V$ is a function of $r$, the slope at that point ($r=2$) is $16$.

However, we can also express this as $\mathrm d V = \mathrm 16 \ dr$. What this tells us is when $r$ changes by a small amount, $V$ changes by $16$ times that amount. As $\Delta r$ approaches $0$, by $\text{slope} = \frac{ \text{rise} }{ \text{run} }$, the slope will approach $16$ at that point. But since $\Delta r$ is a fixed value greater than $0$, this will only be an approximation.

To approximate $\Delta V$ we can just directly substitute: $\Delta r = 0.1 \Rightarrow \mathrm d V = 16 \times 0.1 = 1.6$ .

Toby Mak
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  • Thanks for your response. This is what I was originally thinking, but I updated my question to give more information as to why I don't think this actually makes practical sense. If you're interested, please let me know what you think. – Sean Feb 08 '20 at 06:12