# Margin of error for a tangent line approximation problem related to the volume of a sphere

For my Calculus assignment, I was given this problem:

1. If a right triangle has legs $6$ and $8$, its hypotenuse is $10$. The triangle will be inscribed within a circle with area $25\pi$. (The hypotenuse will be the diameter of the circle).

A. Suppose one leg of the triangle is known to be exactly $6$, but the other leg is known to be $8$ with an error of $\pm h$. What are $x$, $f(x)$ and $a$ in this problem? (Here $x$, $f(x)$, and $a$, relate to the format that was taught for solving Tangent Line Approximation problems; see below.)

C. Now consider the sphere that just contains the triangle (so the hypotenuse is the diameter of the sphere). Use a tangent line approximation to estimate the volume of this sphere.

I ended up with an answer of $\frac{500\pi}{3}\pm 100\pi h$. My teacher said the margin of error was only $40\pi h$, but I don't quite understand why.

The Tangent Line Approximation form we're given is $$f(x) + f'(x)(a - x),$$ where $x$ is the known value, $f'(x)$ is the derivative of $f(x)$, and $a$ is the value being solved for.

For the volume of a sphere, I have $V = \frac{4}{3}\pi r^3$, and for the radius of the sphere I used $\frac{\sqrt{x^2+36}}{2}$.

I found the volume of a triangle with the side measuring $8$ to be $\frac{500 \pi}{3}$ and its derivative $100 \pi$.

Plugging everything into the formula, $$\frac{500 \pi}{3}+100 \pi(8\pm h-8) = \frac{500 \pi}{3} \pm 100 \pi h.$$

I see that if the derivative was $40\pi$, then the margin of error would be what my teacher had said, but I don't see how it makes sense otherwise, or why the derivative would be $40\pi$.

-

For the sphere problem, if $x$ is the length of the side which is nearly $8$, the volume $V(x)$ of the sphere is $\dfrac{4\pi}{3}r^3$, where $r(x)=\dfrac{\sqrt{36+x^2}}{2}$.

Substituting in the formula for the volume of the sphere, and simplifying a bit, we obtain $$V(x)=\frac{\pi}{6}(36+x^2)^{3/2}.$$ We want to use the tangent line approximation to approximate $V(x)$ when $x$ is near $8$. Differentiate. We get $$V'(x)=\pi\frac{x}{2}(36+x^2)^{1/2}.$$ Here the Chain Rule was used.) Set $x=8$. At $x=8$, the derivative is $40\pi$.

The tangent line approximation says that $$V(a)\approx V(8)+40\pi(x-8).$$ If we know that "$x=8\pm h$," where $h$ is positive and small, then by the tangent line approximation we have approximated $V(8)$ within roughly $\pm 40\pi h$.

The area problem is technically easier, since the area is given by the simple expression $\dfrac{\pi}{4}\left(6^2+x^2\right)$.

-
Where's pi in the volume formula, and is the equation with r(x) supposed to be over two? Sorry, I am confused. – mowwwalker Dec 8 '12 at 7:09
Sorry, lost my $\pi$'s. I have a habit of leaving out such constants and putting them in at the end. Except forgot. Thanks. I think they are all back. – André Nicolas Dec 8 '12 at 7:15
I mixed up my a's and x's in the formula - what you posted is, if I remember correctly, what was taught. Also, in your formula for radius, it needs to be over 2, but right now the 2 is just to the right of the equation. – mowwwalker Dec 8 '12 at 7:35
@Walkerneo: I am relieved! With the opposite notation there would have been much confusion in subsequent courses. Have removed comment. Have fixed TeX error that misplaced the $2$. Hope that there are no more typos. – André Nicolas Dec 8 '12 at 7:42

The problem I had was that I was finding the derivative of the volume function, V, in terms of the triangle's side, x, but didn't apply the chain rule to the function I used for the radius, r.

The derivative I had for the volume was $4\pi r^2$, but it should have been $4\pi r^2 r'$

The derivative of $r$ would be $\frac{2}{5}$, which, when multiplied by the derivative I did get ($100\pi$) would have landed me with $40\pi$.

-