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

In summary: Sorry for that.In summary, the ball is thrown up at a speed of 48ft/s from a cliff 432ft above the ground and it says the acceleration is -32. However, the antiderivative is -\cot{x}+c.
  • #1
sjaguar13
49
0
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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  • #2
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]
 
  • #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?
 
  • #4
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]
 
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  • #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]
 
  • #6
As I wrote, cotangens IS the multiplicative inverse of tangens.
Both answers are equally valid.
 
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  • #7
arildno said:
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.
 
  • #8
Arildno,just a little more careful with the terminology,because cotangens IS NOT the inverse of tangense,but arcus tangent is...:wink:

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

Daniel.
 
  • #9
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..:redface:
 
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Related to Positive or Negative Acceleration? and Antiderivative of 1/(sin^2)?

1. What is positive acceleration?

Positive acceleration refers to an increase in the velocity of an object over time. This means that the object is moving faster and faster in a certain direction.

2. What is negative acceleration?

Negative acceleration, also known as deceleration, refers to a decrease in the velocity of an object over time. This means that the object is moving slower and slower in a certain direction.

3. How is acceleration calculated?

Acceleration is calculated by dividing the change in velocity by the change in time. The formula for acceleration is: a = (vf - vi)/t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

4. What is the antiderivative of 1/(sin^2)?

The antiderivative of 1/(sin^2) is -cot(x) + C, where C is a constant and cot(x) is the cotangent function.

5. How is the antiderivative of 1/(sin^2) derived?

The antiderivative of 1/(sin^2) can be derived by using the trigonometric identity 1/(sin^2) = 1 + cot^2(x), and then integrating using the techniques of substitution and integration by parts.

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