i need help on this question: A path is determined by the equations x=1-(5/t^3) and y=2-(1/t^2) for t0. For what time does the line tangent to this path have a slope of 8?

Question

i need help on this question: A path is determined by the equations x=1-(5/t^3) and y=2-(1/t^2) for t>0. For what time does the line tangent to this path have a slope of 8?

owlfred · Answer

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nikvist · Answer

$$x=1-5t^{-3}\quad\Rightarrow\quad dx=15t^{-4}$$
$$y=2-t^{-2}\quad\Rightarrow\quad dy=2t^{-3}$$
$$\frac{dy}{dx}=\frac{2t^{-3}}{15t^{-4}}=\frac{2t}{15}=8\quad\Rightarrow\quad t=60$$