Find out the tangent line(s) to the parametric curve specified by

X = t5 - 4t3

Y = t2

At (0,4)

Solution

Note that there is actually the potential for more than one tangent line here! We will come across into this more after we're completed with the instance.

Firstly we should do is find the derivative thus we can get the slope of the tangent line.

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

= 2t / 5t⁴ - 12t²

= 2 / 5t³ - 12t        

At this point we have got a little problem. The derivative is in terms of t and all we have acquire is an x-y coordinate pair. After that the next step is to determine the value(s) of t that will give this point. We find out these by plugging the x and y values into the parametric equations and by solving for t.

0 = t⁵ - 4t³ = t³ (t² - 4)              ⇒ t = 0, + 2

4 = t2                                             ⇒ t = + 2

It does not matter any value of t which appears in both lists will provide the point. Thus, as there are two values of t that provide the point actually we will get two tangent lines. Though, before we do that let's in fact get the tangent lines.

t = -2

As we already know the x and y-coordinates of the point all that we need to do is find out the slope of the tangent line.

M = dy/dx |_t=-2

= - 1/8

The tangent line that is at t  = -2 is then,

Y = 4 - (1/8) x

t = 2

Once again, all we need is the slope.

M = dy/dx |_t=2

= 1/8  

The tangent line that is at t = 2 is then,

Y = 4 + (1/8) x