# Calculus Homework Question?

A piston is operated by rod [AP] attached to a flywheel of radius 1m. AP=2m. P has coordinates (cos(t), sin(t)) and point A is (-x, 0). t is the angle.

a) Show that x =(-cos(t)) + the square root of (4-sin^2(t))

I'm sorry, there is also a diagram, but I can't show it here. Essentially, there is a triangle in which A is one point connected to point P(cos(t), sin(t)) by the hypotenuse [AP], which equals 2m. Then x is another side length connecting the origin to point A. Point P is connected to the origin and is essentially 1m long. I would really appreciate any ideas for this! We are studying trig derivatives right now. I thought this could be solved by finding x using the cosine rule, but my method did not show the correct answer. Thank you very much!

### 1 Answer

- RahilLv 57 years agoBest Answer
This is just a case of Pythagoras' theorem.

Draw a vertical line from P down to the x-axis (which is the extension of line AO). Let us call the point where the line drawn down from P hits the x-axis the point B.

Triangle ABP is a right angled triangle.

The hypotenuse is AP = 2m, the height is BP = sin(t), the width is AB = AO+OB = x+cos(t)

By Pythagoras we have

4 = sin^2(t)+(x+cos(t))^2

which rearranges to

x = -cos(t) + sqrt(4 - sin^2(t))

### Edit ###

If you want you can also do it by the cosine rule on triangle AOP.

|AP|^2 = |AO|^2 + |OP|^2 -2|AO||OP| cos (angle at O)

4 = x^2 + 1 - 2x cos(180 - t)

x^2 + 2x cos(t) - 3 = 0

Solving the quadratic equation gives us

x = (-2 cos(t) +- sqrt(4 cos^2(t) +12 ))/2

= -cos(t) +- sqrt(cos^2(t) + 3)

Since x is positive we take the positive square root

x = -cos(t) + sqrt(cos^2(t) + 3)

= -cos(t) + sqrt(4 - sin^2(t)) (using cos^2(t) = 1 - sin^2(t))