If we drop a ball from a 100 meter building, what would the speed of the ball be right before it hits the ground? Assume g = 9.81 m/s^2 and no air resistance.
a(t) = -g = -9.8 m/s^2 v(t) = integral (a) dt = -9.8t + v0 p(t) = integral (v) dt = -4.9t^2 + v0t + p0 v^2(final) = v^2(initial) +2*a*(changein y) For this problem, we could use a variation of the first three formulas to find out how long it takes the ball to fall and then plug i into the velocity equation. However, we could just use the last equation as solve this problem with one step: initial velocity = 0, acceleration = -g, changein y = -100 m. v^2(final) = 0 + 2 (-9.81 m/s^2) (-100m) = 1960 m^2/s^2 v (final) = +- 44.3 m/s. The ball's speed, right before it hits the ground, is 44.3 m/s. In my coordinate system (down being negative and up being positive), its velocity would be -44.3 m/s.
What are all asymptotes or removable discontinuities for the following function: f(x) = [x^3 + x^2 - x -1] / [x^2 +x -2]?
Factoring both the numerator (top) and denominator (bottom) we get, f(x) = [ (x+1)^2 * (x-1) ] / [ (x-1) * (x+2) ]. From here, we find a REMOVABLE DISCONTINUITY at the point x = 1. Canceling out x-1 from the top and bottom, we get f(x) = [ x^2 + 2x + 1 ] / [x+2]. This gives us two asymptotes: at x = -2, and a slant asymptote. The slant asymptote can be found by doing long division (or synthetic division) on f(x). After dividing we get the function f(x) = x + 1/(x+2). Therefore, the slant asymptote is y =x. (The remainder 1/(x+2) nears 0 as x nears infinity) Removable discontinuity at x = 2, Vertical asymptote at x = -2, slant asymptote at y = x.
"Area and Volume" Question: What is the volume of the solid generated by rotating the region defined by y = x^2 +2 , x=0, and x=2 about the line y=0?
The radius of this solid at any point "x" is defined as x^2 +2 for 0 <= x <= 2. (I wish I could put a diagram here but..) To visualize what we'll do: think of this solid as an infinite amount of circles stacked on top of each other. The top circle is when x=0, and the bottom circle is when x=2. At x=0, the circle has a radius of 2 [(0)^2 +2]. At x=2, the circle has a radius of 6 [(2)^2+2]. The area for a circle is pi * (radius)^2. Therefore, the area for any circle is pi*(x^2+2)^2 for 0<=x<=2. To find this volume, we will "add up" the areas for the infinite amount of circles with an integral! So, in general form: integral pi*r^2 dx. In this case it is (integral sign from 0 to 2) [pi * (x^2 +2)^2 dx]. Solving this, we arrive at the answer of 376/15 units^3. [I didn't show work here, since student should know how to solve integrals before learning this topic.]