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

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Discussion Overview

The discussion revolves around two main questions: the nature of acceleration when a ball is thrown upward and the antiderivative of a specific mathematical expression involving trigonometric functions. The scope includes conceptual understanding of physics and calculus.

Discussion Character

  • Conceptual clarification
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants suggest that the acceleration of a ball thrown upward is negative due to the force of gravity acting downward, regardless of the initial direction of the throw.
  • Others question the interpretation of acceleration, prompting a consideration of the forces acting on the ball after it has been thrown.
  • Regarding the antiderivative, one participant initially proposes an expression but expresses uncertainty about the correctness of the term involving tangent.
  • Another participant corrects the antiderivative, suggesting that the term should involve cotangent instead of tangent, leading to a revised expression.
  • Some participants note that both expressions involving tangent and cotangent can be seen as valid, though they emphasize the importance of precise terminology in mathematical contexts.
  • There is a discussion about the definitions of cotangent and tangent, with some participants clarifying that cotangent is the reciprocal of tangent, while others point out that "inverse" has a specific meaning in mathematics.

Areas of Agreement / Disagreement

Participants generally agree that the acceleration due to gravity is negative, but there is some debate over the terminology used to describe the relationship between tangent and cotangent. The discussion on the antiderivative remains somewhat unresolved, with different expressions being proposed and validated by different participants.

Contextual Notes

There are nuances in the definitions of trigonometric functions that may lead to confusion, particularly regarding the terms "inverse" and "reciprocal." The discussion also reflects varying levels of certainty about the correctness of mathematical expressions.

sjaguar13
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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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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]
 
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?
 
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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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]
 
As I wrote, cotangens IS the multiplicative inverse of tangens.
Both answers are equally valid.
 
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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.
 
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.
 
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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