Estefano's question at Yahoo! Answers involving a linear approximation

MarkFL

Staff member
Here is the question:

Need help with calculus word problem?

Please explain how to solve it.
An ancient counterfeiter clipped the edges of a gold coin. The coin was originally a cylinder with a radius of 10 mm and a thickness of 2 mm, and the counterfeiter stripped 0.1 mm from around the edges. Approximate the amount of gold stripped from this coin to the nearest mm3.
Here is a link to the question:

Need help with calculus word problem? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

MarkFL

Staff member
Hello Estefano,

For problems like this, I like to begin with the approximation:

$$\displaystyle \frac{\Delta V}{\Delta r}\approx\frac{dV}{dr}$$

$$\displaystyle \Delta V\approx\frac{dV}{dr}\cdot\Delta r$$

$$\displaystyle V(r+\Delta r)-V(r)\approx\frac{dV}{dr}\cdot\Delta r$$

Now, with:

$$\displaystyle V=\pi r^2h\,\therefore\,\frac{dV}{dr}=2\pi rh$$

and letting $$\displaystyle r=9.9\text{ mm},\,\Delta r=0.1\text{ mm}$$ we find:

$$\displaystyle V(10)-V(9.9)\approx\left(2\pi(9.9)(2) \right)\cdot(0.1)=3.96\pi\text{ mm}^3$$

For comparison, the exact value of the change in volume is:

$$\displaystyle V(r+\Delta r)-V(r)=\pi(10)^2(2)-\pi(9.9)^2(2)=2\pi(1.99)=3.98\pi\text{ mm}^3$$

Because of rounding to the nearest unit, the linear approximate gives us a change of volume of $12\text{ mm}^3$, whereas the exact value would be rounded to $13\text{ mm}^3$.

To Estefano and any other guests viewing this topic, I invite and encourage you to register and post other calculus problems in our Calculus forum.

Best Regards,

Mark.