(a)[3]
Show that $\frac{dy}{dx} = e^{-2t}$ is true.
(b)[3]
Hence show that the normal to the curve, where $t = -1$, goes through the point $\left(0, 3 - \frac{1}{e^4}\right)$.
Mathematics 9709 · AS & A Level · Differentiation
Differentiation — practice question
The curve is specified by the parametric equations $x = te^{2t}$ and $y = t^2 + t + 3$.
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This 6-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Obtain $\dfrac{dx}{dt}=e^{2t}+2te^{2t}$ by differentiating $x$.” …
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