Question #b5ac6

Question

1436 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-02T18:08:34

Normal would be vertical line parallel to y axis, hence its slope would be undefined.

$\frac{d x}{d t} = - a sin t$ $dx/dt= -a sin t$ and $\frac{d y}{d t} = a cos t$ $dy/dt= a cos t$

this gives $\frac{d y}{d x} = - \frac{cos t}{sin t}$ $dy/dx = - cost /sint$

At t= $90^{o} = \frac{π}{2}$ $90^o= pi/2$ , $\frac{d y}{d x} = - \frac{0}{1} = 0$ $dy/dx= -0/1=0$

The slope at t= pi/2 is thus 0, that means the tangent is a horizontal line that is parallel to x axis.

The normal to this horizontal tangent would thus be a vertical line. Its slope would therefor be undefined,