kdgg0909gn

2021-12-06

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

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

Robert Harris

Beginner2021-12-07Added 23 answers

Step 1

The given parametric equations are:

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

To find the second derivative of y with respect to x, i.e.$\frac{dy}{dx}$

Differentiate the given y and x with respect to t.

That is,

$\frac{dx}{dt}=\frac{d}{dt}\left(\frac{1}{{t}^{2}}\right)$

1)$\frac{dx}{dt}=-\frac{2}{{t}^{3}}$

$\frac{dy}{dt}=\frac{d}{dt}({t}^{2}-4t+1)$

2)$\frac{dy}{dt}=2t-4$

Step 2

From equations (1) and (2).

$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2t-4}{-\frac{2}{{t}^{3}}}$

$\frac{dy}{dx}=-{t}^{3}(t-2)$

The given parametric equations are:

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

Differentiate the given y and x with respect to t.

That is,

1)

2)

Step 2

From equations (1) and (2).

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