caculus derivative, differentials?

Question

1. use differentials to approxmiate rad(4.9)

2. for what value of x is the line tangent to y =x^2 parallel to the line tangent to y = x^(1/2)
(i know you have to get the derivative of y for both equation and put it equal to each other, but i cant figure it out..

3. A 17foot ladder is sliding down a wall at a rate of -5ft/sec. When the top of the ladder is 8 ft from the gound how fast is the foot of the ladder sliding away from the wall.

Answer 1

1. Unfortunately, I'm not sure what you mean by rad.

2. Your explanation is correct. For y1 = x^2, y1' = 2x. For y2 = sqrt(x), y2' = 1 / (2*sqrt(x)). Now set them equal to each other. 2x = 1 / (2*sqrt(x) ==> 4x*sqrt(x) = 1 ==> 16x^3 = 1 ==> x = (1/16)^(1/3) = 0.397, or (1 / 2^(4/3)) if you want to express it precisely.

3. If the top of the ladder is y feet above the ground, let us say that the foot is x feet from the wall. By the Pythagorean theorem, we know that x^2 + y^2 = 17^2 = 289. So we can say x = sqrt(289 - y^2) and therefore dx/dy = (1 / (2*sqrt(289 - y^2)))(-2y) by the chain rule. Simply evaluate the expression for y = 8.