Question

Given the curve x=\cos(t) + 2\cos(2t) ; y=\sin(t) + 2\sin(2t), graph the curve to discover where it crosses itself and find equations of both tangents at that point.

Answer 1

Answer and Explanation:

First the graph.



From it, we see that the curve crosses itself at the point (−2,0). We need the derivatives of the parametric equations. We have

dxdt=−sint−4sin2t

dydt=cost+4cos2t

Then

dydx=dy/dtdx/dt=cost+4cos2t−sint−4sin2t

Unfortunately, this is not good. We will want to express the trig functions explicitly at the point (−2,0). Using the y-coordinate we get

0=sint+2sin2t=sint+2(2sintcost)=sint(1+4cost)

1+4cost=0

4cost=−1

cost=−14

Now, since cos2t+sin2t=1, we must have

(−14)2+sin2t=1

sin2t=1−116=1516

sint=±√154

Since we approach the point from the second quadrant, we have sint=√154. So we also have

sin2t=2sintcost=2(√154)(−14)=−√158

and

cos2t=2cos2t−1=2(−14)2−1=216−1=−78

Then

dydx=dy/dtdx/dt=cost+4cos2t−sint−4sin2t=(−14)+4(−78)−(√154)−4(−√158)=−15/4√15/4=−15√15=−√15

Now, since we have the point (−2,0), we get

y=mx+b

0=−√15(−2)+b

b=−2√15

The equation for this tangent line is

y=−√15x−2√15

For the other tangent line, everything is upside down, i.e. it's exactly the same except for the signs:

y=√15x+2√15