How do you differentiate the following parametric equation: $x (t) = t^{3} - 5 t, y (t) = (t - 3)$ $x(t)=t^3-5t, y(t)=(t-3)$ ?

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How do you differentiate the following parametric equation: $x (t) = t^{3} - 5 t, y (t) = (t - 3)$ $x(t)=t^3-5t, y(t)=(t-3)$ ?

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Answer 1 · 2016-01-13T15:45:01

$\frac{d y}{d x} = \frac{1}{3 t^{2} - 5}$ $dy/dx = 1/(3t^2 - 5 )$

Explanation:

$x = t^{3} - 5 t \Rightarrow \frac{d x}{d t} = 3 t^{2} - 5$ $x = t^3 - 5t rArr dx/dt = 3t^2 - 5$

and $y = t - 3 \Rightarrow \frac{d y}{d t} = 1$ $y = t - 3 rArr dy/dt = 1$

$\frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x}$ $dy/dx = dy/dt xx dt/dx$

note that: $\frac{d t}{d x} = \frac{1}{\frac{d x}{d t}}$ $dt/dx =( 1)/(dx/dt)$

$\Rightarrow \frac{d y}{d x} = \frac{1}{3 t^{2} - 5}$ $rArr dy/dx = 1/(3t^2 - 5 )$

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How do you differentiate the following parametric equation: $x (t) = t^{3} - 5 t, y (t) = (t - 3)$ $x(t)=t^3-5t, y(t)=(t-3)$ ?

1 Answer

Explanation:

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How do you differentiate the following parametric equation: x(t)=t^3-5t, y(t)=(t-3) x(t)=t3−5t,y(t)=(t−3) x(t)=t^3-5t, y(t)=(t-3) ?

1 Answer

Explanation:

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How do you differentiate the following parametric equation: $x (t) = t^{3} - 5 t, y (t) = (t - 3)$ $x(t)=t^3-5t, y(t)=(t-3)$ ?