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

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Solution

$Here, x = t^{2} + 3 t - 8 and y = 2 t^{2} - 2 t - 5 \frac{d x}{d t} = 2 t + 3 and \frac{d y}{d t} = 4 t - 2 ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{4 t - 2}{2 t + 3} Now, Slope of the tangent= {(\frac{d y}{d x})}_{t = 2} = \frac{8 - 2}{4 + 3} = \frac{6}{7}$

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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 ?

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. 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.

Here,x=t2+3t-8 and y=2t2-2t-5dxdt=2t+3 and dydt=4t-2∴dydx=dydtdxdt=4t-22t+3Now,Slope of the tangent=dydxt=2=8-24+3=67

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

$Here, x = t^{2} + 3 t - 8 and y = 2 t^{2} - 2 t - 5 \frac{d x}{d t} = 2 t + 3 and \frac{d y}{d t} = 4 t - 2 ∴ \frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{4 t - 2}{2 t + 3} Now, Slope of the tangent= {(\frac{d y}{d x})}_{t = 2} = \frac{8 - 2}{4 + 3} = \frac{6}{7}$