Recent content by Exeneva

Sorry, that's what I meant, the second derivative (just forgot the squared on the bottom). My main issue is I'm not sure how I was supposed to have found the second derivative. I tried to do a reverse chain rule but I am pretty sure I did it wrong.

I think I tried to do a chain rule but screwed up somewhere.

Homework Statement x = t - e^{t} y = t + e^{-t} Find dy/dx and d^{2}y/dx.Homework Equations Derivative equations.The Attempt at a Solution dy/dt = 1 - e^{-t} dx/dt = 1 - e^{t} The dy/dx I came up with is: dy/dx = (1 - e^{-t}) / (1 - e^{t}) Second derivative I came up with is: d^{2}y/dx = -...

Homework Statement Write the polar equation for the graph y = x. Homework Equations x = r cos \theta y = r sin \theta The Attempt at a Solution I came up with \theta = \pi/4 because at \pi/4, the x and y coordinates match each other. I'm not sure this is correct, though.

Homework Statement r = 3 sin \vartheta 0 \leq \vartheta \leq \pi/3 Homework Equations Arc Length: \int \sqrt{r^{2} + (dr/d\vartheta)^{2}}d\vartheta The Attempt at a Solution r^{2} = 9 (sin \vartheta)^{2} = 9 (1/2 - cos 2\vartheta/2) r^{2} = 9/2 - 9/2 cos 2\vartheta...