Question

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

Answer 1

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}

₌ \((\frac{4t-2}{2t+3})_{at\,t = 2}\)

= \(\frac{4(2)-2}{2(2)+3}\) 

= \(\frac{6}{7}\)