Question

If x= 5t-t^3, y=t^2+4t find dy/dx at t= 3

Answer 1

Answer:

$\boxed{\mathfrak{\frac{dy}{dx} = -2.2}}$

Given:

$\sf x = 5t - {t}^{3} \\ \sf y = {t}^{2} + 4t$

To Find:

$\sf \frac{dy}{dx} \: at \: t = 3$

Explanation:

$\sf \implies \frac{dx}{dy} \\ \\ \sf \implies \frac{dx}{dy} \times \frac{dt}{dt} \\ \\ \sf \implies \frac{ \frac{dx}{dt} }{ \frac{dy}{dt} } \\ \\ \sf \implies \frac{ \frac{d}{dt} (5t - {t}^{3}) }{ \frac{d}{dt}( {t}^{2} + 4t) } \\ \\ \sf \implies \frac{5 - 3 {t}^{2} }{2t + 4} \\ \\ \sf At \: t = 3 : \\ \sf \implies \frac{5 - 3 {(3)}^{2} }{2(3) + 4} \\ \\ \sf \implies \frac{5 - 27}{6 + 4} \\ \\ \sf \implies - \frac{22}{10} \\ \\ \sf \implies - 2.2$