Question

Find the arc length of the curve on the given interval. $$ x=t^{2}, \quad y=2 t \quad 0 \leq t \leq 2 $$

Short answer

The arc length of the curve on the given interval is \( \frac{2}{3}(5\sqrt{5}-2)\).

Step-by-step solution

Find the derivatives

First, find the derivative of both \(x(t)\) and \(y(t)\). Derivative of \(x(t)\) is \(x'(t)=2t\) and the derivative of \(y(t)\) is \(y'(t)=2\).

Calculate the square of each derivative

Next, find the squares of each derivative. The square of \(x'(t)\) is \((x'(t))^2 = (2t)^2 = 4t^2\), and the square of \(y'(t)\) is \((y'(t))^2 = 2^2 = 4\).

Find the sum of the squares

Add up the squares of the derivatives. The sum of the squares is \(4t^2 + 4\).

Calculate the square root of the sum

Next step is taking the square root of the sum of squares, resulting in \(\sqrt{4t^2 + 4}\).

Set up the integral

Then, set up the integral, which extends from the given parameter range of 0 to 2: \(\int_0^2 \sqrt{4t^2 + 4} dt\).

Simplify the integral

We can simplify this to \(\int_0^2 2\sqrt{t^2 + 1} dt\).

Resolve the integral

Finally, resolve the integral. Here, a substitution can be used. If we let \(u=t^2+1\), then \(du=2t dt\). The integral can be rewritten to \(\int_1^{5} \sqrt{u} du\). Calculate this final integral to obtain the result.

Calculate the definite integral

Calculate the definite integral, which equals \( \frac{2}{3}[u^{3/2}]_1^5 = \frac{2}{3}(5\sqrt{5}-2)\).