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

(C) Given the parametric equations of the curve: $x = t^2 + 3t - 8$ and $y = 2t^2 - 2t - 5$. First,find the derivatives with respect to $t$: $\frac{dx

Vedclass · Accepted Answer

$6/7$

Vedclass · Answer

(C) Given the parametric equations of the curve: $x = t^2 + 3t - 8$ and $y = 2t^2 - 2t - 5$. First,find the derivatives with respect to $t$: $\frac{dx}{dt} = 2t + 3$ $\frac{dy}{dt} = 4t - 2$ The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{4t - 2}{2t + 3}$. To find the value of $t$ at the point $(2, -1)$,substitute the coordinates into the equations: $2 = t^2 + 3t - 8 \Rightarrow t^2 + 3t - 10 = 0 \Rightarrow (t+5)(t-2) = 0 \Rightarrow t = 2, -5$. $-1 = 2t^2 - 2t - 5 \Rightarrow 2t^2 - 2t - 4 = 0 \Rightarrow t^2 - t - 2 = 0 \Rightarrow (t-2)(t+1) = 0 \Rightarrow t = 2, -1$. The common value is $t = 2$. Now,substitute $t = 2$ into the slope formula: $\frac{dy}{dx} = \frac{4(2) - 2}{2(2) + 3} = \frac{8 - 2}{4 + 3} = \frac{6}{7}$.

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

Similar Questions

For a positive constant $a$,find $\frac{dy}{dx}$,where $y = a^{t+\frac{1}{t}}$ and $x = \left(t+\frac{1}{t}\right)^{a}$.

If $\sin y = \sin 3t$ and $x = \sin t$, then $\frac{dy}{dx} = $

If $x = a \cos^3 \theta$ and $y = a \sin^3 \theta$,then find the value of $\sqrt{1 + \left(\frac{dy}{dx}\right)^2}$.

If $x=a(\cos t+t \sin t)$ and $y=a(\sin t-t \cos t),$ find $\frac{d^{2} y}{d x^{2}}.$

The angle made by the tangent at $\theta = \pi / 3$ on the curve $x = a(\theta + \sin \theta)$,$y = a(1 - \cos \theta)$ with the $X$-axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module