How do you find $\frac{d y}{d x}$ $dy/dx$ for the curve $x = t \cdot sin (t)$ $x=t*sin(t)$ , $y = t^{2} + 2$ $y=t^2+2$ ?

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Answer 1 · 2014-08-28T02:11:36

To find the derivative of a parametric function, you use the formula:

$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$ $dy/dx = (dy/dt)/(dx/dt)$ , which is a rearranged form of the chain rule.

To use this, we must first derive $y$ $y$ and $x$ $x$ separately, then place the result of $\frac{d y}{d t}$ $dy/dt$ over $\frac{d x}{d t}$ $dx/dt$ .

$y = t^{2} + 2$ $y=t^2 + 2$

$\frac{d y}{d t} = 2 t$ $dy/dt = 2t$ (Power Rule)

$x = t sin (t)$ $x=tsin(t)$

$\frac{d x}{d t} = sin (t) + t cos (t)$ $dx/dt = sin(t) + tcos(t)$ (Product Rule)

Placing these into our formula for the derivative of parametric equations, we have:

$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{2 t}{sin (t) + t cos (t)}$ $dy/dx = (dy/dt)/(dx/dt) = (2t)/(sin(t)+tcos(t))$

How do you find dy/dxdydxdy/dx for the curve x=t*sin(t)x=t⋅sin(t)x=t*sin(t), y=t^2+2y=t2+2y=t^2+2 ?