1. Linear Approximation Application

I'm having some trouble applying linear approximation and changes, so I'm stuck on this question.

The diameter of a tree = 10 in.
Circumference increases by 2 in
About how much did the tree's diameter grow? How much did the tree's cross section area grow?

I get the concept of linearization, but I have no idea how to apply it in questions like this. Thanks again for your time.

2. Originally Posted by Wyau
I'm having some trouble applying linear approximation and changes, so I'm stuck on this question.

The diameter of a tree = 10 in.
Circumference increases by 2 in
About how much did the tree's diameter grow? How much did the tree's cross section area grow?

I get the concept of linearization, but I have no idea how to apply it in questions like this. Thanks again for your time.
The linear approximation to a change in the argument of a function f(x) is
$\displaystyle f(x + \Delta x) \approx f(x) + f^{\prime}(x) \cdot \Delta x$

So we know that the initial diameter of the tree is d = 10 in. And we know that the circumference of the tree grows by 2 in.

The formula for the circumference is
$\displaystyle C = \pi d$
so
$\displaystyle d(C) = \frac{C}{\pi}$

I'm assuming you have to use the linear approximation equation here (though the answer turns out to be exact, not an approximation) so
$\displaystyle C = \pi d = \pi \cdot 10~in = 31.415~in$
and
$\displaystyle d^{\prime}(C) = \frac{1}{\pi}$

Thus
$\displaystyle d(C + \Delta C) = \frac{C}{\pi} + \frac{1}{\pi} \cdot \Delta C$

$\displaystyle d(31.415~in + 2~in) = \frac{31.415~in}{\pi} + \frac{1}{\pi} \cdot 2~in = 10.637~in$

For the second problem, we know that the cross-section is a circle and the area of a circle is:
$\displaystyle A = \pi r^2 = \pi \left ( \frac{d}{2} \right )^2 = \frac{\pi}{4}d^2$

So
$\displaystyle A^{\prime}(d) = \frac{\pi}{2}d$

The change in d is:
$\displaystyle \Delta d = 10.637~in - 10~in = 0.637~in$

So
$\displaystyle A(d + \Delta d) \approx A(d) + A^{\prime}(d) \cdot \Delta d$

$\displaystyle \approx \frac{\pi}{4}(10~in)^2 + \frac{\pi}{2}(10~in) \cdot (0.637~in)$

$\displaystyle \approx 88.540 ~ in^2$

(The actual area is $\displaystyle A = 88.8581~in^2$, so this is a good approximation.)

-Dan