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

• sjaguar13
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.
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.

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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 $$\frac{1}{\sin^{2}x}$$

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?

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

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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 $$-\cot{x}+c$$

$$\frac{2}{5}x^{5/2}-\cot{x}+c$$
and not

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

As I wrote, cotangens IS the multiplicative inverse of tangens.

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arildno said:
As I wrote, cotangens IS the inverse of tangens.

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.

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.

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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## 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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