Question:
Find the slope of the tangent to the curve $x=t^{2}+3 t-8, y=2 t^{2}-2 t-5$ at $t=2$.
Solution:
Given that $x=t^{2}+3 t-8, y=2 t^{2}-2 t-5$
$\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=2 \mathrm{t}+3, \frac{\mathrm{dy}}{\mathrm{dt}}=4 \mathrm{t}-2$
$\therefore \frac{d y}{d x}=\frac{4 t-2}{2 t+3}$
Now,
Slope of the tangent $($ at $t=2)=\frac{8-2}{4+3}=\frac{6}{7}$
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