Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$x=\cos t+\cos 2t$ and $y = \sin t+\sin 2t$ at $(-1,1)$

I did dy/dt and dx/dt and then dy/dx and got $\displaystyle \frac{\cos t+2\cos 2t}{-\sin t-2\sin 2t}$

Now I'm confused, do I plug in the -1 and 1 to get two answers and then subtract them from each other to the get a final slope? I need to get the slope so that I can make an equation in the y=mx+b form.

share|cite|improve this question
up vote 1 down vote accepted


Find the value(s) of $t$ which give the point $(-1,1)$.

Note however, that by inspection, the point $(-1,1)$ on the curve corresponds to $t=\frac{\pi}{2}$ in the parametric equations. So you find the slope at $t=\frac{\pi}{2}.$

share|cite|improve this answer
so i set my x and y equations equal to -1 and 1 respectively? – Ryan Jul 18 '11 at 5:00
yes Ryan. That's what you'd normally do. All though we can do this by inspection as well. I'll add to my answer shortly. – Nana Jul 18 '11 at 5:03

To find the value(s) of t that yield (-1, 1), I computed $x^2+y^2$ and got (modulo my usual errors) $2(1+cos^3 t)$. Setting this equal to 2 gives $\cos(t) = 0$. This gives two possible values of t, only one of which gives the desired values of x and y. As to why I did this, I noticed that the sin/cos terms with args of t and 2t would sum to 2 (in $x^2+y^2$), so I thought I would see what happened to the cross terms. Turns out it helped.

share|cite|improve this answer
Good idea. But you do not mean $2(1+\cos^3 t)$. For one thing that would give $\cos t=-1$, and your $\cos t=0$ is right. – André Nicolas Jul 18 '11 at 5:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.