Recent content by frosty8688

For some reason I get 2.82 seconds for t.

1. At t=0, a stone is dropped from the top of a cliff above a lake. Another stone is thrown downward 1.6 s later from the same point with an initial speed of 32 m/s. Both stones hit the water at the same instant. Find the height of the cliff. 2. d_{1}=\frac{1}{2}gt_{1}^{2}...

Which is: 2\int(sec^{2}θ-1)dθ = 2(tanθ-θ)+C

I see what you mean. Ok, here is what I have after the dx substitution: \int\frac{\sqrt{x^{2}-4}}{x}dx = \int\frac{2tanθ}{2secθ}2secθtanθdθ = 2\int tan^{2}θdθ

1. ∫\frac{\sqrt{x^{2}-4}}{x} dx, x=2secθ, dx=2secθtanθ dθ 2. \sqrt{x^{2}-a^{2}},sec^{2}θ-1=tan^{2}θ 3...

I have \frac{1}{3}(2\sqrt{2}-1)a^{3}

I got it when a is evaluated x = +/- 1 and since x=a +/- 1^2 + +/- 1^2 = 2 and when the lower limit is evaluated x = 0, but a > 0 so that makes the expression equal to 1.

Is there an equation I can use to compute the values of x?

So \left|x\right| = a and -a.

So we let a = 1 and 1+1 = 2 and it would be sqrt 2^3 which would be 2 sqrt 2 - (0+1)^3 = 1 all to a^3 * 1/3.

1. \int^{a}_{0} x\sqrt{x^{2}+a^{2}} a > 0 2. u = x^{2} + a^{2}, du = 2x 3. \frac{1}{2}\int^{a}_{0}\frac{2u^{3/2}}{3} = \frac{1}{3}\int ^{a}_{0}(x^{2}+a^{2})^{3/2} How do I solve this? Any hints?

If I simplify it, here's how it looks, \frac{20\sqrt{3}}{9+4\sqrt{3}} and if I multiply by 2, I get\frac{40\sqrt{3}}{9+4\sqrt{3}}. Does this look right?

1. A piece of wire 10 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral triangle. How should the wire be cut so that the total area enclosed is (a) a maximum? (b) A minimum? 2. A_{s}(x) = x^{2}, A_{t}(x)=\frac{\sqrt{3}}{4}x^{2}...

Here is how I got that answer \frac{\frac{-2x^{3}}{2(1-x^{2})^{3/2}}}+\frac{2x}{2\sqrt{1-x^{2}}}-\frac{2x}{\sqrt{1-x^{2}}}{x^{2}}-2\frac{\frac{-x^{2}}{\sqrt{1-x^{2}}}}-\sqrt{1-x^{2}}{x^{3}} =...

Here is what I have \frac{\frac{-x}{\sqrt{1-x^{2}}}-\frac{x^{3}}{(1-x^{2})^{3/2}}} x^{2} + \frac{2}{x^{3}} The squared on the top is supposed to go with the x on the bottom.