Recent content by hell0

Good question. Honestly that's all the info I have. Nice professor I have, huh? Anyway I made the sides of the triangle be 1 and 1, and found the hypotenuse to be sqrt(2). Then used the theorem again but this time found x(1) via [x(1)]^2+(1/2)^2=sqrt(2)^2 since y(1) is given in the problem and...

1. Suppose we have ladder laying against a wall, with x sub 0 = 1 and y sub 0 = 1. Given that y(1) = 1 and x′(1) = 1, find y'(1). Okay so using Pythagorean's Theorem x^2 + y^2 = L, I found that the remaining side is sqrt(2). After taking the derivative I got 2x(t)x'(t) + 2y(t)y'(t) = 0. The...