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

 
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Apr15-05, 04:21 AM   #1
 

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


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 [tex]\frac{x^2}{\sqrt{x}} + \frac{1}{sin^2x}[/tex]
I can got:
[tex]\frac{2}{5}x^{5/2}+\frac{1}{\tan{x}}+c[/tex]

But I am pretty sure [tex]\frac{1}{\tan{x}}[/tex] isn't right.
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Apr15-05, 04:39 AM   #2
 
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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, [tex]\frac{d}{dx}\frac{1}{tan(x)}=-\frac{1}{\sin^{2}x}[/tex] , not [tex]\frac{1}{\sin^{2}x}[/tex]
Apr15-05, 04:46 AM   #3
 
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:
[tex]\frac{2}{5}x^{5/2}-\frac{1}{\tan{x}}+c[/tex]
The only thing wrong was the sign?
Apr15-05, 04:50 AM   #4
 
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Positive or Negative Acceleration? and Antiderivative of 1/(sin^2)?


Yes, right on both problems.
Note that you will most often find the multiplicative inverse of the tangens function called cotangens:
[tex]cot(x)\equiv\frac{1}{tan(x)}[/tex]
Apr15-05, 05:21 AM   #5
 
I just thought of something else. [tex]If \csc{x} = \frac{1}{\sin{x}} , then \csc^{2}x = \frac{1}{\sin^{2}x}[/tex]

The antiderivative of [tex]\csc^{2}{x}[/tex] is [tex]-\cot{x}+c[/tex]

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

[tex]\frac{2}{5}x^{5/2}-\frac{1}{\tan{x}}+c[/tex]
Apr15-05, 05:24 AM   #6
 
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As I wrote, cotangens IS the multiplicative inverse of tangens.
Both answers are equally valid.
Apr15-05, 09:30 AM   #7
 
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Quote by arildno
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, with out any other explanation, you are saying f(g(x))= g(f(x))= x.
Apr15-05, 09:32 AM   #8
 
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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.
Apr15-05, 09:53 AM   #9
 
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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..
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