Question

Find the equation of the curve that passes through the point (x, y) = (0, 0) and has an arc length on the interval x is between 0 and pi over 4 inclusive given by the integral the integral from 0 to pi over 4 of the square root of the quantity 1 plus the cosine squared x, dx . (5 points)

Answer 1

Answer:

$\displaystyle y=\sin(x),\text{ where } 0\leq x\leq\pi/4$

Step-by-step explanation:

The curve passes through the point (x, y) = (0, 0) and has an arc length on the interval [0, π/4] given by the integral:

$\displaystyle \int_{0}^{\pi/4}\sqrt{1+\cos^2(x)}\, dx$

And we want to find the equation of the curve.

Recall that arc length is given by:

$\displaystyle L=\int_a^b\sqrt{1+\Big(\frac{dy}{dx}\Big)^2}\, dx$

Rewrite our original integral:

$\displaystyle \int_{0}^{\pi/4}\sqrt{1+(\cos(x))^2}\, dx$

So:

$\displaystyle \frac{dy}{dx}=\cos(x)$

It follows that:

$\displaystyle y=\sin(x)+C$

Using the initial condition:

$0=\sin(0)+C\Rightarrow 0=0+C\Rightarrow C=0$

The equation for our curve is:

$\displaystyle y=\sin(x),\text{ where } 0\leq x\leq\pi/4$