(D) The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.Given $x = t^2 + 3t - 8$,we have $\frac{dx}{dt} = 2t + 3$.Given $y = 2t

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(D) The slope of the tangent is given by $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.Given $x = t^2 + 3t - 8$,we have $\frac{dx}{dt} = 2t + 3$.Given $y = 2t^2 - 2t - 5$,we have $\frac{dy}{dt} = 4t - 2$.Therefore,$\frac{dy}{dx} = \frac{4t - 2}{2t + 3}$.At $t = 2$,the slope is $\frac{dy}{dx} = \frac{4(2) - 2}{2(2) + 3} = \frac{8 - 2}{4 + 3} = \frac{6}{7}$.

Class 12 Mathematics Continuity and Differentiation Derivatives of Functions in Parametric Forms

Easy

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

A
$11$
B
$13$
C
$9$
D
$6/7$

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Derivatives of Functions in Parametric Forms

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

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