Area with Parametric Equations — Question 4

Question 4 – Solution

Solution

1. The endpoints correspond to $t=0$ and $t=2$. The chord through $(0,0)$ and $(4,8)$ has slope $$m=\frac{8-0}{4-0}=2,$$ so its equation is $y=2x$.

2. Along the parametric curve, $x=t^2$ and $y=t^3$. The chord’s height at the same $x$-coordinate is $$y_{\mathrm{chord}}=2x=2t^2.$$ For $0\le t\le2$, $2t^2-t^3=t^2(2-t)\ge0$, so the chord lies above the curve.

3. Since $dx/dt=2t$, the area between them is $$A=\int_0^2\bigl(y_{\mathrm{chord}}-y_{\mathrm{curve}}\bigr)\,dx =\int_0^2(2t^2-t^3)(2t)\,dt.$$

4. Evaluate: $$\begin{aligned} A&=\int_0^2(4t^3-2t^4)\,dt\\ &=\left[t^4-\frac25t^5\right]_0^2 =16-\frac{64}{5} =\boxed{\frac{16}{5}}. \end{aligned}$$