Linear Approximation Application

**Wyau** · September 4th 2007, 09:44 PM

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.

**topsquark** · September 5th 2007, 05:58 AM

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
$f(x + \Delta x) \approx f(x) + f^{\prime}(x) \cdot \Delta x$

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
$C = \pi d$
so
$d(C) = \frac{C}{\pi}$

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

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

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

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

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

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

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

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

$\approx 88.540 ~ in^2$

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

-Dan

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