The slope of the tangent at (1, 2) to the cur | vedclass

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

Question

(A) Given the parametric equations of the curve are $x = t^2 - 7t + 7$ and $y = t^2 - 4t - 10$. To find the value of $t$ at the point $(1, 2)$, we set

Vedclass · Accepted Answer

$\frac{8}{5}$

Vedclass · Answer

(A) Given the parametric equations of the curve are $x = t^2 - 7t + 7$ and $y = t^2 - 4t - 10$. To find the value of $t$ at the point $(1, 2)$, we set $x = 1$ and $y = 2$. For $x = 1$: $t^2 - 7t + 7 = 1 \Rightarrow t^2 - 7t + 6 = 0 \Rightarrow (t - 6)(t - 1) = 0$, so $t = 1$ or $t = 6$. For $y = 2$: $t^2 - 4t - 10 = 2 \Rightarrow t^2 - 4t - 12 = 0 \Rightarrow (t - 6)(t + 2) = 0$, so $t = 6$ or $t = -2$. The common value is $t = 6$. Now, we find the derivatives with respect to $t$: $\frac{dx}{dt} = 2t - 7$ and $\frac{dy}{dt} = 2t - 4$. At $t = 6$: $\frac{dx}{dt} = 2(6) - 7 = 12 - 7 = 5$. $\frac{dy}{dt} = 2(6) - 4 = 12 - 4 = 8$. The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{8}{5}$.

The slope of the tangent at $(1, 2)$ to the curve $x = t^2 - 7t + 7$ and $y = t^2 - 4t - 10$ is:

Similar Questions

Find the equations of the tangent and normal to the given curve at the indicated point: $x = \cos t, y = \sin t$ at $t = \frac{\pi}{4}$.

If $x=a(t+\sin t)$ and $y=a(1-\cos t)$,then $\frac{dy}{dx}=$

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

If $x = at^2$ and $y = 2at$,then $\frac{dy}{dx} = $ . . . . . . .

If $\frac{dy}{dx} = 4$ and $\frac{d^2y}{dx^2} = -3$ at a point $P$ on the curve $y = f(x)$, then $\left(\frac{d^2x}{dy^2}\right)_P = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module