Question

A wheel of radius 1 rolls along a straight line, say the r-axis. A point P is located halfway between the center of the wheel and the rim; assume P starts at the point (0, 1/2). As the wheel rolls, P traces a curve; find parametric equations for the curve. Erercises 10.5. 2. Consider the curve of exercise 6 in section 10.4. Find the area under one arch of the curve. Final answer is 9?/4

Answer 1

Answer:

Part a: The parametric equations of the curve are as indicated $x=\frac{1}{2}(\theta-sin \theta)\\y=\frac{1}{2}+\frac{1}{2}(1-cos \theta)$

Part b: The area under the curve is $\frac{9 \pi}{4}$

Step-by-step explanation:

Part a

As the wheel rolls the path traced by the point is a cycloid which is as given as

$x=\frac{r}{2}(\theta-sin \theta)\\y=\frac{1}{2}+\frac{r}{2}(1-cos \theta)$

As the radius is 1 the equation is

$x=\frac{1}{2}(\theta-sin \theta)\\y=\frac{1}{2}+\frac{1}{2}(1-cos \theta)$

The parametric equations of the curve are as indicated above.

Part b

The area under the curve is given as

$\int_{0}^{2 \pi} ydx$

Here

y is given as

$y=\frac{1}{2}+\frac{1}{2}(1-cos \theta)$

and x is given as

$x=\frac{1}{2}(\theta-sin \theta)$

Finding its differential as

${dx}=[\frac{1}{2}-cos \theta]}d\theta$

Substituting in the equation and solving the equation

$\int_{0}^{2 \pi} ydx\\\int_{0}^{2 \pi} [\frac{1}{2}+\frac{1}{2}(1-cos \theta)][\frac{1}{2}-cos \theta]}d\theta]\\\int_{0}^{2 \pi} [\frac{(cos(\theta)-2)(2cos(\theta)-1)}{4}d\theta]\\=\frac{9 \pi}{4}$

So the area under the curve is $\frac{9 \pi}{4}$