Question

if x= sin(t) and y=cos(t) can someone tell me how to find the second derivative?

Answer 1

\[ x = \sin (t)\]
\[\frac{\mathrm{d}x}{\mathrm{d}t}=\cos(t)\]
\[\frac{\mathrm{d}^2x}{\mathrm{d}t^2}=-\sin(t)\]

Then the same goes for y

The derivative of sine is cosine, and the derivative of cosine is negative sine.

^^

Answer 2

And then, again, the derivative of negative cosine is positive sine.

Answer 3

but when you do dy/dt all over dx/dt and you put the dy over the dx then you get - sin t/cos t and that equals -tan t but the derivative of -tan t isn't in the answer choices :(

Answer 4

You're looking for
\[\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\]
?

Answer 5

yep

Answer 6

I don't understand the question, then. Both y and x are in terms of t, so dy/dx = 0

If you're just stacking them on top of each other, then it wouldn't be tangent, it would be cotangent.

What are your choices?