Question

Show that the curvex = 2 cos t, y = 3 sin t cos t has two tangents at (0, 0) and find their equations.

Answer 1

This is the case of parametric derivatives in which x and y are expressed as functions of a variable t.

The first derivative implies that:


`(dy)/(dx) = ( (dy)/(dt) )/( (dx)/(dt) )`

So we need to find both derivatives as functions of the variable t, then:


`(dy)/(dt) = 3[cos(t)cos(t)+sin(t)(-sin(t))]`

`(dy)/(dt) = 3[cos^(2) (t)-sin^(2) (t)]`


`(dx)/(dt) = -2sin(t)`

Thus:


`(dy)/(dx) = (-3[cos^(2) (t)-sin^(2) (t)])/(2sin(t))`

At (0,0)
`x=0` and
`y=0`, so:


`\left \{ {{0=2cos(t)} \atop {0=3sin(t)cos(t)}} \right.`
∴
`\left \{ {{cos(t)=0} \atop {sin(t)cos(t)=0}} \right.`

There are two values between -π and π which satisfy these equations simultaneously, namely:

`t =` π/2

`t =` -π/2

The equation of a straight line given a point and its slope is

`y - y_(0) = m(x- x_(0) )`

Given that the point is (0,0), then the equation can be written as:

`y = mx`

So we will find two straight lines:

`\left \{ {{y= m_(1)x } \atop {y= m_(2)x }} \right.`

For
`t = \pi /2`:


`m_(1)= (dy)/(dx) = (-3[cos^(2) ( \pi /2)-sin^(2) ( \pi /2)])/(2sin( \pi /2)) = (3)/(2)`

For
`t = -\pi /2`:


`m_(2)= (dy)/(dx) = (-3[cos^(2) ( -\pi /2)-sin^(2) ( -\pi /2)])/(2sin(- \pi /2)) = -(3)/(2)`

Lastly:

`\left \{ {{y= (3)/(2) x } \atop {y= -(3)/(2) x }} \right.`