# Find the slope of the tangent to the curve x = t^2 + 3t – 8, y = 2t^2 – 2t – 5 at t = 2.

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

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Slope of the tangent =$\frac{dy}{dx}$

$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$\frac{4t-2}{2t+3}$

$\frac{dy}{dx}$ at t = 2

$(\frac{4t-2}{2t+3})_{at\,t = 2}$

$\frac{4(2)-2}{2(2)+3}$

$\frac{6}{7}$