# Positive or Negative Acceleration? and Antiderivative of 1/(sin^2)?

1. Apr 15, 2005

### sjaguar13

I have two quick questions. The first is, if a ball is thrown from a bridge 25 meter above the ground at 49m/s, is the acceleration negative? I thought since it was thrown up, it would be positive, but the book has a similar problem, a ball is thrown upward at a speed of 48ft/s from a cliff 432ft above the ground, and it says the acceleration is -32.

Second, what is the antiderivative of $$\frac{x^2}{\sqrt{x}} + \frac{1}{sin^2x}$$
I can got:
$$\frac{2}{5}x^{5/2}+\frac{1}{\tan{x}}+c$$

But I am pretty sure $$\frac{1}{\tan{x}}$$ isn't right.

Last edited: Apr 15, 2005
2. Apr 15, 2005

### arildno

As for the acceleration question:
1. WHAT FORCES ACT UPON THE BALL ONCE IT HAS LEFT YOUR HAND?
What is that force's direction? (Assuming there is only one force acting on the ball)
2. How is a force related to acceleration?

As for the derivative:
You're right, $$\frac{d}{dx}\frac{1}{tan(x)}=-\frac{1}{\sin^{2}x}$$ , not $$\frac{1}{\sin^{2}x}$$

3. Apr 15, 2005

### sjaguar13

Gravity will always pull it down, no matter if it was thrown up or down, so acceleration is always negative (assumming it's on Earth, under normal conditions)?

The antiderivative is:
$$\frac{2}{5}x^{5/2}-\frac{1}{\tan{x}}+c$$
The only thing wrong was the sign?

4. Apr 15, 2005

### arildno

Yes, right on both problems.
Note that you will most often find the multiplicative inverse of the tangens function called cotangens:
$$cot(x)\equiv\frac{1}{tan(x)}$$

Last edited: Apr 15, 2005
5. Apr 15, 2005

### sjaguar13

I just thought of something else. $$If \csc{x} = \frac{1}{\sin{x}} , then \csc^{2}x = \frac{1}{\sin^{2}x}$$

The antiderivative of $$\csc^{2}{x}$$ is $$-\cot{x}+c$$

Making the final answer be:
$$\frac{2}{5}x^{5/2}-\cot{x}+c$$
and not

$$\frac{2}{5}x^{5/2}-\frac{1}{\tan{x}}+c$$

6. Apr 15, 2005

### arildno

As I wrote, cotangens IS the multiplicative inverse of tangens.
Both answers are equally valid.

Last edited: Apr 15, 2005
7. Apr 15, 2005

### HallsofIvy

No, it is not. Cotangent is the reciprocal or multiplicative inverse of tangent. If you say a function f is the "inverse" of a function g, with out any other explanation, you are saying f(g(x))= g(f(x))= x.

8. Apr 15, 2005

### dextercioby

Arildno,just a little more careful with the terminology,because cotangens IS NOT the inverse of tangense,but arcus tangent is...

Cotangent is cos of 'x' divided by'sin of 'x' by definition.

Daniel.

9. Apr 15, 2005

### arildno

Oh dear, oh dear..
"multiplicative" is such a long word it is tempting to dispense with it, but:
Guess I'll have to be more careful with not using lazy short-cuts in the future..

Last edited: Apr 15, 2005