# Math Questions -- Limits

Discussion in 'Physics & Math' started by Amadeusboy, Sep 17, 2008.

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22
In my review of calculus I have finally made it to the concept of the limit. I was working some of the problems and worked one problem which had an answer which did not seem intuitively correct. The question is Problem Twelve from page sixty of Boyce and DiPrima's Calculus.

The surface area $S$ of a sphere is related to the radius $r$ by $S=4 \pi r^2$.

(a) Find the average rate at which $S$ changes as $r$ changes from $r_o$ to $r_o +h$.

(b) Find the instantaneous rate at which $S$ changes with $r$ when $r=r_o$.

(c) Find the rate at which $S$ changes with $r$ when $r=2$.​

Part (a) was solved as follows:

$\Large\begin{eqnarray} \Delta S_{av}(r_o)&=&\frac{S(r_o+h)-S(r_o)}{r_o+h-r_o}\\&=&\frac{4\pi (r_o+h)^2-4\pi r_o^2}{h}\\&=&\frac{4\pi r_o^2 +8\pi r_oh+4\pi h^2-4\pi r_o^2}{h}\\&=&8\pi r_o+4\pi h.\end{eqnarray}$

Part (b) was solved as follows:

$\begin{eqnarray} \Delta S_{inst}(r_o)&=&\lim_{r\to r_o}\Delta S_{av}\\&=&\lim_{r\to r_o}8\pi r_o+4\pi h\\&=&8\pi r_o.\end{eqnarray}$

Finally, part (c) was solved as follows:

$\Delta S_{inst}(r)=8\pi r$;

hence,

$\Delta S_{inst}(2)=16\pi.$

Are these the correct answers? Should the average and instantaneous rate at which $S$ changes with $r$ be based on multiples of $\pi$? If one doubles the radius does one not quadruple the surface area? Or is this statement the ratio of two results and the interpretation that $16\pi$ is the slope of the line tangent to $S(r)=4\pi r^2$ at $r=2$ the correct one?

Also, I don't know if the the symbols $\Delta S_{av}(r_o) \tex{ and } \Delta S_{inst}(r_o)$ are the correct symbols to use for these expressions.

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5. ### CheskiChipsBannedBanned

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Don't question the math. Never do that.

7. ### rpennerFully WiredValued Senior Member

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Doubling the linear measure always quadruples the surface area for a similar figure.

${{\Delta_h S(r_0)}\over{\Delta_h r_0}} = {{S(r_0 + h) - S(r_0)}\over{r_0 + h - r_0}} = 4 \pi{{\left( r_0 + h \right)^2 - r^2_0}\over{h}} = 4 \pi {{r_0^2 + 2 r_0 h + h^2 - r^2_0}\over{h}} = 4 \pi \left( 2 r_0 + h \right) = 8 \pi r_0 + 4 \pi h$

$S' (r_0) = \lim_{h \to 0} {{\Delta_h S(r_0)}\over{h}} = \lim_{h \to 0} 8 \pi r_0 + 4 \pi h = 8 \pi r_0$

$S' (2) = \left. S'(r_0) \right| _{r_0 = 2} = \left. 8 \pi r_0 \right| _{r_0 = 2} = 8 \pi \times 2 = 16 \pi$

So other than quibbles with notation for average speed, it all looks good to me.

As I factored out the 4 pi early on, it is obvious that 4pi must survive to the end in the expressions for average and instantaneous rate. The instantanous rate is the slope of the tangent line. The way I wrote it, the average rate is the slope of a line which passes through y = S(x) at two points an its slope is just ${{\Delta y} \over{\Delta x}} = {{\Delta y} \over{h}} = {{S(x+h) - S(x)} \over{h}}$.

Last edited: Sep 17, 2008
8. ### §outh§taris feeling causticRegistered Senior Member

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4,832
Oh ok, I was looking at the derivative when I asked that. Never mind!

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Here are other problems about which I had questions. The questions are adapted from numbers thirty-two and thirty-three from page seventy-two of the previously mentioned text.

Given the definition of a limit as:
Let $f$ be defined at each point in some open interval containing the point $c$, except possibly at $c$ itself. The the limit of $f(x)$ as $x$approaches $c$ is said to be $L$, written $\lim_{x\to c}f(x)=L\text{,}$ if for each number $\epsilon>0$, there exists a corresponding number $\delta >0$ such that if $0<\middle| x-c\middle|<\delta$ then $0<\middle| f(x)-L\middle|<\epsilon$.

32c. Let $f(x)=x^2$ with $c=2$ and$L=3$. Find $\delta(\epsilon)$ such that if $0<\middle| x-c\middle|<\delta$ then $0<\middle| f(x)-L\middle|<\epsilon$. Hint: First restrict $\delta$ so that $\delta<1$.

We want $\delta$ such that if $0<\middle| x-2\middle|<\delta$, then $0<\middle| x^2-4\middle|<\epsilon$.

First, rewrite $0<\middle| x^2-4\middle|<\epsilon$ as $0<\middle| x-2\middle|\middle| x+2\middle|<\epsilon$ and choose the open interval $0<\middle| x-2\middle|<\delta$ to be the interval $0<\middle| x-2\middle|<\delta<1$. This restricts $\delta$ as the hint suggests.

Then $-1<x-2<1 \Rightarrow 3<x+2<5$, which implies $\middle| x+2\middle| <5$.

Now, $\middle| x-2\middle|<\delta$ and $\middle| x+2\middle|<5$ imply $\middle| x-2\middle|\middle|x+2\middle|<5\delta$, but $\middle| x-2\middle|\middle|x+2\middle|=\middle|x^2-4\middle|<5\delta$

Thus we want $5\delta<\epsilon$, or $\delta<\frac{\epsilon}{5}$ and we want the minimum of $(1,\frac{\epsilon}{5})$ (1 is our original choice of $\delta$). Hence, $\delta < \tex{min}(1,\frac{\epsilon}{5})$.

33c Let $f(x)=\frac{1}{x}$ with $c=2$ and$L=\frac{1}{2}$. Find $\delta(\epsilon)$ such that if $0<\middle| x-c\middle|<\delta$ then $0<\middle| f(x)-L\middle|<\epsilon$. Hint: First restrict $\delta$ so that $\delta<1$.

Again, we want $\delta$ such that if $0<\middle| x-2\middle|<\delta$, then $0<\middle| \frac{1}{x}-\frac{1}{2}\middle|<\epsilon$.

First, choose the open interval $0<\middle| x-2\middle|<\delta$ to be the interval $0<\middle| x-2\middle|<\delta<1$. This restricts $\delta$ as the hint suggests.

Then $-1<x-2<1 \Rightarrow 1<x<3$, which implies both $\frac{1}{x}$$<1$ and $\middle |\frac{1}{x}\middle |$$<1$

Now, $\middle| \frac{1}{x}-\frac{1}{2}\middle|=\middle|\frac{1}{2x}\middle| \middle|2-x \middle|=\middle| \frac{1}{2x} \middle| \middle|x-2\middle|\middle|-1\middle| =\middle| \frac{1}{2x} \middle| \middle|x-2\middle|<\epsilon \Rightarrow \middle| \frac{1}{2} \middle|\middle| \frac{1}{x} \middle| \middle|x-2\middle|\text{ }= \text{ }\frac{1}{2} \middle| \frac{1}{x} \middle| \middle|x-2\middle|<\epsilon$.

Since $\middle| \frac{1}{x} \middle|$ $<1$ and $\middle|x-2 \middle|<\delta$, $\frac{1}{2} \middle| \frac{1}{x} \middle| \middle|x-2\middle|\text{ }<\text{ }\frac{1}{2} \middle|x-2\middle| <\frac{1}{2}\delta$. Thus, we want $\frac{\delta}{2}<\epsilon$ or, after taking the above restriction into consideration $\delta<\text{min}(1,2\epsilon)$.

It has been awhile since I have done these types of problems and I'm wondering are these solutions sound?

Last edited: Aug 20, 2009
10. ### paulfrRegistered Senior Member

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227
Why does the answer to part b drop the 4 pi h ?

Lim x ==> x0 [ f(x) + k ] = g(x0)

and

Lim x ==> x0 [ k ] = k

So shouldn't the k appear in the answer ?

Last edited: Aug 20, 2009
11. ### CptBorkValued Senior Member

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6,355
I checked your answer and I didn't see anything wrong with it. You should be a bit more careful with your notation. Basically, you've shown that $\Delta S\approx 8\pi r \Delta r$, with the approximate equality becoming exact as $\Delta S$ and $\Delta r$ approach infinitesimally small values. You can't ask the question "what happens if I double the radius" and use this equation to find the answer, because it only holds for infinitesimally small changes in radius. The surface area does indeed quadruple as radius doubles, because actual surface area is still described by the original equation, $S=4\pi r^2$.

Remember that when you have a function like $y=Ax$ or $y=Ax^2$ and find the slope of the tangent as you have (what is known in calculus as "taking the derivative"), the final answer will still be proportional to $A$. That's what's happened here, there's a factor of $\pi$ appearing in the original formula, so when you find its slope, that slope will still be proportional to $\pi$.

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The first step in the solution to the neglected to mention the difference quotient from which I started.

$\Large\begin{eqnarray} \Delta S_{av}&=&\frac{S(r)-S(r_o)}{r-r_o}.\\ \Delta S_{inst}(r)&=&\lim_{r\to r_o}\Delta S_{av}\\& =&\lim_{r\to r_o}\frac{S(r)-S(r_o)}{r-r_o}\\& =&4\pi\lim_{r\to r_o}\frac{r^2-r_o^2}{r-r_o}\\& =&4\pi\lim_{r\to r_o}(r+r_o)\\& =&8\pi r_o. \end{eqnarray}$

Sorry for the confusion this error caused.

Does anyone see any errors in the sixth post.

Last edited: Aug 21, 2009