The slope of the tangent to the curve x = t^2 | vedclass

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

Question

(D) Given $x = t^2 + 3t - 8 = 2$ $t^2 + 3t - 10 = 0$ $(t + 5)(t - 2) = 0$ $t = -5$ or $t = 2$ Also $y = 2t^2 - 2t - 5 = -1$ $2t^2 - 2t - 4 = 0$ $t^2 -

Vedclass · Accepted Answer

$6/7$

Vedclass · Answer

(D) Given $x = t^2 + 3t - 8 = 2$ $t^2 + 3t - 10 = 0$ $(t + 5)(t - 2) = 0$ $t = -5$ or $t = 2$ Also $y = 2t^2 - 2t - 5 = -1$ $2t^2 - 2t - 4 = 0$ $t^2 - t - 2 = 0$ $(t - 2)(t + 1) = 0$ $t = 2$ or $t = -1$ Since the point $(2, -1)$ must satisfy both equations,we take the common value $t = 2$. The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$. $\frac{dx}{dt} = 2t + 3$ $\frac{dy}{dt} = 4t - 2$ $\frac{dy}{dx} = \frac{4t - 2}{2t + 3}$ At $t = 2$: $\frac{dy}{dx} = \frac{4(2) - 2}{2(2) + 3} = \frac{8 - 2}{4 + 3} = \frac{6}{7}$.

The slope of the tangent to the curve $x = t^2 + 3t - 8$ and $y = 2t^2 - 2t - 5$ at the point $(2, -1)$ is:

Similar Questions

Let $y = t^{10} + 1$ and $x = t^8 + 1,$ then $\frac{d^2y}{dx^2}$ is

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x = a \sec \theta$ and $y = b \tan \theta$.

Find the equation of the tangent to the curve given by $x = a \sin^{3} t$ and $y = b \cos^{3} t$ at the point where $t = \frac{\pi}{2}$.

If $x=e^{t}(\sin t-\cos t)$ and $y=e^{t}(\sin t+\cos t)$,then $\frac{dy}{dx}$ at $t=\frac{\pi}{3}$ is

Derivative of $\sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right)$ with respect to $\cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module