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

The derivative of $\frac{1-x^2}{1+x^2}$ with respect to $\frac{2x}{1+x^2}$ at $x=2$ is

The derivative of $f(\tan x)$ with respect to $g(\sec x)$ at $x = \frac{\pi}{4}$,where $f^{\prime}(1) = 2$ and $g^{\prime}(\sqrt{2}) = 4$,is:

If $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$,then $(x^2 + 4) \left(\frac{dy}{dx}\right)^2 =$

If $x = \log t$ and $y + 1 = \frac{1}{t}$,then $e^{-x} \frac{d^{2} x}{d y^{2}} + \frac{d x}{d y} = $

The equation of the normal to the curve $x = \theta + \sin \theta, y = 1 + \cos \theta$ at $\theta = \frac{\pi}{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module