Solved: "Find the second derivative of y with respect..."

kdgg0909gn

Answered question

2021-12-06

Find the second derivative of y with respect to x from the parametric equations given.

x = \frac{1}{t^{2}}, y = t^{2} - 4 t + 1

Robert Harris

Beginner2021-12-07Added 23 answers

Step 1
The given parametric equations are:

x = \frac{1}{t^{2}}, y = t^{2} - 4 t + 1

To find the second derivative of y with respect to x, i.e.

\frac{d y}{d x}

Differentiate the given y and x with respect to t.
That is,

\frac{d x}{d t} = \frac{d}{d t} (\frac{1}{t^{2}})

\frac{d x}{d t} = - \frac{2}{t^{3}}

\frac{d y}{d t} = \frac{d}{d t} (t^{2} - 4 t + 1)

\frac{d y}{d t} = 2 t - 4

Step 2
From equations (1) and (2).

\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{2 t - 4}{- \frac{2}{t^{3}}}

\frac{d y}{d x} = - t^{3} (t - 2)

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