Answer to Question #115179 in Calculus for Bibhuti bhusan dehury

Question

Find the length of the curve given by x=t^3, y=2t^2 in 0<=t<=1. What is the slope of the curve at t=1/2.

Answer 1

Given equation of of the curve in parametric form is $x=t^3, \: y=2t^2$ . Now, we have to find the length of this curve from $0 \leq t \leq 1$ .

Since, length of a parametric curve is given by,

$$L=\int_{t_1}^{t_2} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt \hspace{1.5cm} (\star)$$

Thus, plugin

$$\frac{dx}{dt}=3t^2 \: \& \frac{dy}{dt}=4t$$

in equation $(\star)$, we get

$$L= \int_{0}^{1} \sqrt{(3t^2)^2 + (4t)^2} dt$$

$$\implies L= \int_{0}^{1} t\sqrt{9t^2 + 16} dt$$

Let, substitute $u= 9t^2 +16 \implies \frac{du}{18t}=dt$ and $\def\arraystretch{1.5} \begin{array}{c:c:c} u & 16 & 25\\ \hline t & 0 & 1 \\ \end{array}$ , hence

$$L=\frac{1}{18}\int_{16}^{25} \sqrt{u} \:du\\ L=\frac{1}{18}\frac{ u^{\frac{3}{2}}}{\frac{3}{2}} \bigg|_{16}^{25} \\L=\frac{1}{27}u^{\frac{3}{2}}\bigg|_{16}^{25}\\L=\frac{1}{27}(25^{\frac{3}{2}}-16^{\frac{3}{2}})=\frac{61}{27}$$

Therefore, the required length is $\frac{61}{27}$ .

Now we know that if $y=f(x)$ is a parametric curve such that $x=x(t) \&y=y(t)$ ,then slope of that curve at a point $P$ with given parameter $t=t_0$ is given by,

$$y'=\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\bigg|_{t=t_0}\hspace{1.5cm} (\spades)$$

Thus, slope of the given curve at $t=\frac{1}{2}$ is given by $\frac{dy}{dx}$ ,thus from equation$(\spades)$ we get,

$$\frac{dy}{dx}=\frac{4}{3t} |_{t=\frac{1}{2}}=\frac{8}{3}$$

Hence, we are done.