Find the slope of the tangent to the curve $x = t^{2} + 3 t - 8, y = 2 t^{2} - 2 t - 5$ at $t = 2$ .

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Solution

Given $x = t^{2} + 3 t - 8, y = 2 t^{2} - 2 t - 5$

Slope of curve at t=2,

$\frac{d y}{d x}$

$= \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

$= \frac{\frac{d (2 t^{2} - 2 t - 5)}{d t}}{\frac{d (t^{2} + 3 t - 8)}{d t}}$

$= \frac{4 t - 2}{2 t + 3}$

as $t = 2$

$= \frac{6}{7}$

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Similar questions

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

Q. What is the slope of the tangent to the curve

x = t^{2} + 3 t - 8, y = 2 t^{2} - 2 t - 5

at t = 2 ?

Q. The slope of tangent to the curve

x = t^{2} + 3 t - 8, y = 2 t^{2} - 2 t - 5

at the point

(2, - 1)

is :

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

The slope of the tangent to the curve x = t² + 3t − 8, y = 2t² − 2t − 5 at the point (2, −1) is

(a) $\frac{22}{7}$
(b) $\frac{6}{7}$
(c) $\frac{7}{6}$
(d) $- \frac{6}{7}$

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Find the slope of the tangent to the curve x=t2+3t−8,y=2t2−2t−5 at t=2.

Given x=t2+3t−8,y=2t2−2t−5Slope of curve at t=2,dydx=dydtdxdt=d(2t2−2t−5)dtd(t2+3t−8)dt=4t−22t+3as t=2=67

Find the slope of the tangent to the curve $x = t^{2} + 3 t - 8, y = 2 t^{2} - 2 t - 5$ at $t = 2$ .

Given $x = t^{2} + 3 t - 8, y = 2 t^{2} - 2 t - 5$

Slope of curve at t=2,

$\frac{d y}{d x}$

$= \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

$= \frac{\frac{d (2 t^{2} - 2 t - 5)}{d t}}{\frac{d (t^{2} + 3 t - 8)}{d t}}$

$= \frac{4 t - 2}{2 t + 3}$

as $t = 2$

$= \frac{6}{7}$