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miyoko23q3

Answered question

2021-11-30

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

x = 1 - \frac{1}{t}, y = 6 - \frac{7}{t} + \frac{2}{t^{2}}

Befory

Beginner2021-12-01Added 19 answers

Step 1
Given:

x = 1 - \frac{1}{t}, y = 6 - \frac{7}{t} + \frac{2}{t^{2}}

To find:
The second derivative of y with respect to x from the given parametric equations.
Step 2
Consider, the given parametric equations

x = 1 - \frac{1}{t}, y = 6 - \frac{7}{t} + \frac{2}{t^{2}}

The derivative of y w.r.t, x can be defines as,

\frac{d y}{d x} = \frac{(\frac{d y}{d t})}{(\frac{d x}{d t})}

Here,

\frac{d y}{d t} = \frac{d}{d t} (6 - \frac{7}{t} + \frac{2}{t^{2}})

\frac{d y}{d t} = - (- \frac{7}{t^{2}}) + (- \frac{2 \times 2}{t^{2}})

\frac{d y}{d t} = \frac{7}{t^{2}} - \frac{4}{t^{3}}

Step 3
Now,

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

\Rightarrow \frac{d y}{d x} = \frac{(\frac{7}{t^{2}} - \frac{4}{t^{3}})}{(\frac{1}{t^{2}})}

Again differentiate w.r.t,x

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

Step 4
To find the value of

\frac{d^{2} y}{{d t}^{2}}

Differmtiate equation (1) w.r.t,t

\Rightarrow \frac{d^{2} y}{{d t}^{2}} = \frac{d}{}

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