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

(B) Given the parametric equations of the curve are $x = t^2 + 3t - 8$ and $y = 2t^2 - 2t - 5$. First,we find the value of $t$ at the point $(2, -1)$.

Vedclass · Accepted Answer

$6/7$

Vedclass · Answer

(B) Given the parametric equations of the curve are $x = t^2 + 3t - 8$ and $y = 2t^2 - 2t - 5$. First,we find the value of $t$ at the point $(2, -1)$. Substituting $x = 2$ into the equation for $x$: $2 = t^2 + 3t - 8 \implies t^2 + 3t - 10 = 0 \implies (t+5)(t-2) = 0$. Thus,$t = 2$ or $t = -5$. Checking $t = 2$ in the equation for $y$: $y = 2(2)^2 - 2(2) - 5 = 8 - 4 - 5 = -1$. This matches the given point. Now,calculate the derivatives with respect to $t$: $\frac{dx}{dt} = 2t + 3$ and $\frac{dy}{dt} = 4t - 2$. The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t - 2}{2t + 3}$. At $t = 2$,the slope is $\frac{4(2) - 2}{2(2) + 3} = \frac{8 - 2}{4 + 3} = \frac{6}{7}$.

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

Similar Questions

If $\sqrt{y-\sqrt{y-\sqrt{y-\ldots \infty}}} = \sqrt{x+\sqrt{x+\sqrt{x+\ldots \infty}}}$,then $\frac{dy}{dx} = $

If $x = t + \frac{1}{t}$ and $y = t - \frac{1}{t}$,then $\frac{d^2y}{dx^2}$ is equal to

If $x = \sin 2\theta \cos 3\theta$ and $y = \sin 3\theta \cos 2\theta$, then find $\frac{dy}{dx}$.

If $x = a(\cos \theta + \theta \sin \theta)$, $y = f(\theta)$, $f(2\pi) = 0$, $\frac{dy}{dx} = \frac{\tan \theta}{\theta}$, $\theta \neq 0$ and $\theta \neq (2n+1)\frac{\pi}{2}$, then $f\left(\frac{\pi}{3}\right) = $

The derivative of $\sin x$ with respect to $\log x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module