What is the slope of the tangent to the curve | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the parametric equations of the curve are $x = a \sin^3 t$ and $y = a \cos^3 t$.First,we find the derivatives with respect to $t$:$\frac{dx}

Vedclass · Accepted Answer

$-\frac{1}{\sqrt{3}}$

Vedclass · Answer

(D) Given the parametric equations of the curve are $x = a \sin^3 t$ and $y = a \cos^3 t$.First,we find the derivatives with respect to $t$:$\frac{dx}{dt} = 3a \sin^2 t \cos t$$\frac{dy}{dt} = -3a \cos^2 t \sin t$Now,the slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.$\frac{dy}{dx} = \frac{-3a \cos^2 t \sin t}{3a \sin^2 t \cos t} = -\frac{\cos t}{\sin t} = -\cot t$.At $t = \frac{\pi}{3}$,the slope is:$\left( \frac{dy}{dx} \right)_{t = \pi/3} = -\cot\left( \frac{\pi}{3} \right) = -\frac{1}{\sqrt{3}}$.

What is the slope of the tangent to the curve $x = a \sin^3 t, y = a \cos^3 t$ at $t = \frac{\pi}{3}$?

Similar Questions

If $x = \sin \theta, y = \sin^3 \theta$ then $\frac{d^2 y}{d x^2}$ at $\theta = \frac{\pi}{2}$ is . . . . . .

If $f^{\prime}(x)=\sqrt{2 x^2-1}$ and $y=f(x^3)$,then find the value of $\frac{dy}{dx}$ at $x=1$.

If $x=3\left[\sin t-\log \left(\cot \frac{t}{2}\right)\right]$ and $y=6\left[\cos t+\log \left(\tan \frac{t}{2}\right)\right]$,then $\frac{dy}{dx}=$

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x=2at^{2}$ and $y=at^{4}$.

If $x=e^\theta(\sin \theta-\cos \theta)$ and $y=e^\theta(\sin \theta+\cos \theta)$,then $\frac{dy}{dx}$ at $\theta=\frac{\pi}{4}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module