New Notes Blog: `x=3costheta ,y=3sintheta` Find all points (if any) of horizontal and vertical tangency to the curve.

Wednesday, 16 July 2014

$\displaystyle{x}={3}{\cos{\theta}},{y}={3}{\sin{\theta}}$ Find all points (if any) of horizontal and vertical tangency to the curve.

$\displaystyle{x}={3}{\cos{\theta}}$

$\displaystyle{y}={3}{\sin{\theta}}$

First, take the derivative of x and y with respect to $\displaystyle\theta$ .

$\displaystyle\frac{{\left.{d}{x}}}{{{d}\theta}}=-{3}{\sin{\theta}}$

$\displaystyle\frac{{\left.{d}{y}}}{{{d}\theta}}={3}{\cos{\theta}}$

Take note that the slope of a tangent is equal to dy/dx.

$\displaystyle{m}=\frac{{\left.{d}{y}}}{{\left.{d}{x}}}$

To get the dy/dx of a parametric equation, apply the formula:

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{\frac{{\left.{d}{y}}}{{{d}\theta}}}}{{\frac{{\left.{d}{x}}}{{{d}\theta}}}}$

When the tangent line is horizontal, the slope is zero.

$\displaystyle{0}=\frac{{\frac{{\left.{d}{y}}}{{{d}\theta}}}}{{\frac{{\left.{d}{x}}}{{{d}\theta}}}}$

This implies that the graph of...

$\displaystyle{x}={3}{\cos{\theta}}$

$\displaystyle{y}={3}{\sin{\theta}}$

First, take the derivative of x and y with respect to $\displaystyle\theta$ .

$\displaystyle\frac{{\left.{d}{x}}}{{{d}\theta}}=-{3}{\sin{\theta}}$

$\displaystyle\frac{{\left.{d}{y}}}{{{d}\theta}}={3}{\cos{\theta}}$

Take note that the slope of a tangent is equal to dy/dx.

$\displaystyle{m}=\frac{{\left.{d}{y}}}{{\left.{d}{x}}}$

To get the dy/dx of a parametric equation, apply the formula:

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}=\frac{{\frac{{\left.{d}{y}}}{{{d}\theta}}}}{{\frac{{\left.{d}{x}}}{{{d}\theta}}}}$

When the tangent line is horizontal, the slope is zero.

$\displaystyle{0}=\frac{{\frac{{\left.{d}{y}}}{{{d}\theta}}}}{{\frac{{\left.{d}{x}}}{{{d}\theta}}}}$

This implies that the graph of the parametric equation will have a horizontal tangent when $\displaystyle\frac{{\left.{d}{y}}}{{{d}\theta}}={0}$ and $\displaystyle\frac{{\left.{d}{x}}}{{{d}\theta}}\ne{0}$ .

So, set the derivative of y equal to zero.

$\displaystyle\frac{{\left.{d}{y}}}{{{d}\theta}}={0}$

$\displaystyle{3}{\cos{\theta}}={0}$

$\displaystyle{\cos{\theta}}={0}$

$\displaystyle\theta=\frac{\pi}{{2}},\frac{{{3}\pi}}{{2}}$

These are the values of theta in which the graph of parametric equation will have horizontal tangents.

Then, substitute these values to the parametric equation to get the points (x,y).

$\displaystyle\theta=\frac{\pi}{{2}}$

$\displaystyle{x}={3}{\cos{\theta}}={3}{\cos{{\left(\frac{\pi}{{2}}\right)}}}={3}\cdot{0}={0}$

$\displaystyle{y}={3}{\sin{\theta}}={3}{\sin{{\left(\frac{\pi}{{2}}\right)}}}={3}\cdot{1}={3}$

$\displaystyle\theta=\frac{{{3}\pi}}{{2}}$

$\displaystyle{x}={3}{\cos{\theta}}={3}\frac{{\cos{{\left({3}\pi\right)}}}}{{2}}={3}\cdot{0}={0}$

$\displaystyle{y}={3}{\sin{\theta}}={3}\frac{{\sin{{\left({3}\pi\right)}}}}{{2}}={3}\cdot{\left(-{1}\right)}=-{3}$

Therefore, the parametric equation has horizontal tangent at points (0,3) and (0,-3).

Moreover, when the tangent line is vertical, the slope is undefined.

$\displaystyle{u}{n}{d}{e}{f{{i}}}{n}{e}{d}=\frac{{\frac{{\left.{d}{y}}}{{{d}\theta}}}}{{\frac{{\left.{d}{x}}}{{{d}\theta}}}}$

This happens when $\displaystyle\frac{{\left.{d}{x}}}{{{d}\theta}}={0}$ , but $\displaystyle\frac{{\left.{d}{y}}}{{{d}\theta}}\ne{0}$ .

So, set the derivative of x equal to zero.

$\displaystyle\frac{{\left.{d}{x}}}{{{d}\theta}}={0}$

$\displaystyle-{3}{\sin{\theta}}={0}$

$\displaystyle{\sin{\theta}}={0}$

$\displaystyle\theta={0},\pi$

These are the values of theta in which the graph of parametric equation will have vertical tangents.

Then, plug-in these values to the parametric equation to get the points (x,y).

$\displaystyle\theta={0}$

$\displaystyle{x}={3}{\cos{\theta}}={3}{\cos{{0}}}={3}\cdot{1}={3}$

$\displaystyle{y}={3}{\sin{\theta}}={3}{\sin{{0}}}={3}\cdot{0}={0}$

$\displaystyle\theta=\pi$

$\displaystyle{x}={3}{\cos{\theta}}={3}{\cos{\pi}}={3}\cdot{\left(-{1}\right)}=-{3}$

$\displaystyle{y}={3}{\sin{\theta}}={3}{\sin{\pi}}={3}\cdot{0}={0}$

Therefore, the parametric equation has vertical tangent at points (3,0) and (-3,0).

New Notes Blog

Wednesday, 16 July 2014

$\displaystyle{x}={3}{\cos{\theta}},{y}={3}{\sin{\theta}}$ Find all points (if any) of horizontal and vertical tangency to the curve.

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Wednesday, 16 July 2014

Find all points (if any) of horizontal and vertical tangency to the curve.

No comments:

Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

$\displaystyle{x}={3}{\cos{\theta}},{y}={3}{\sin{\theta}}$ Find all points (if any) of horizontal and vertical tangency to the curve.

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?