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

[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.

 

[arildno]

As for the acceleration question:
Ask yourself:
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}

 

[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?

 

[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)}

 

[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

 

[arildno]

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

 

[HallsofIvy]

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

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, without any other explanation, you are saying f(g(x))= g(f(x))= x.

 

[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.

 

[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..