1. ## Linear Approximation Problem?

On page 431 of Physics: Calculus, 2d ed., by Eugene Hecht(Pacific Grove, CA: Brooks/Cole, 2000), in the course of derivingthe formula T=2π√L/g for the period of a dendulum oflength L, the author obtains the equation at=-gsinθfor the tangential acceleration of the bob of the pendulum. He then says, "for small angles, the value of θ in radians isvery nearly the value of sin θ; they differ by less than 2%out to about 20 degrees."

(a) Verify the linear approximation at 0 for the sine function:

sinx ˜ x

(b) Use a graphing device to determine the values of x for whichsin x and x differ by less than 2%. Then verify Hecht'sstatement by converting from radians to degrees.

I understand how to do a, where if you find the equation of the tangent line you should end up with y=x. But can someone please help me with b?

2. For part b just plot them on the same screen and estimate.
Here is one plot of
$f(x)=x-\sin(x) \quad g(x)=.02$

their intersection is where the error is equal to 2%.

sine.pdf

3. wow you can actually use 2% as a data on the graph? I don't exactly understand the logic of that though. So would I have to use the amount between, say, .1 and .49? .1 radians = 5.7 degrees , .49 radians = 28.1 degrees. So 28.1-5.7 = 22 degrees? Or am I completely wrong. Thanks for your help by the way.

4. Originally Posted by RezMan
wow you can actually use 2% as a data on the graph? I don't exactly understand the logic of that though. So would I have to use the amount between, say, .1 and .49? .1 radians = 5.7 degrees , .49 radians = 28.1 degrees. So 28.1-5.7 = 22 degrees? Or am I completely wrong. Thanks for your help by the way.
Here is the logic behind the idea. This distance between

$x$

and

$\sin(x)$

is given by

$|x-\sin(x)|$

but we want

$|x-\sin(x)|< 2\% = .02$

So I just found the solution to this equation graphically by finding the intersection of

$f(x)=|x-\sin(x)| \text{ and } g(x)=.02$

5. wow that makes sense. I got confused by wondering what the 100% would be. So the answer would be .49 radians = 28.1 degrees?

6. Yep that is the approximate answer.

7. thank you so much! Your use of the graph kinda inspires me to represent everything in graph form!