Applying inverse trig function

In summary, the conversation was about a problem involving a ladder leaning against a wall and the bottom of the ladder sliding away from the wall at a constant rate. The question was how fast the angle between the ladder and the wall is changing at a specific point. The conversation discussed using inverse trig functions and related rates to solve the problem. The final answer was 1/4 rad/s.
  • #1
SpicyPepper
20
0
Hey people, first time posting. This isn't really homework, but I want to know how to apply inverse trig functions. So, while looking over some problems in my textbook, I came across the following, which I really have no idea how to do.

Homework Statement


A ladder 10ft long leans against a wall. The bottom of the ladder (on the floor) slides away from the wall at 2ft/sec. How fast is the angle between the ladder & the wall changing when the bottom of the ladder is 6ft from the base of the wall?

Homework Equations



The Attempt at a Solution


Since the ladder is 10ft (hypothenuse), the distance from the ladder to wall is 6ft (opposite), then I can figure the wall is effectively 8ft (adjacent).

I assume the angle [tex]\Theta[/tex] can be figured with any inverse trig function.

I know the answer (1/4 rad/s), but that's not really important. I want to be able to understand how to approach this type of problem.
 
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  • #2
Since the triangle stays in contact with the wall, you can say that [tex]x^{2}+y^{2}=10^{2}[/tex] at all times, and [tex]x=10\cos\theta, y=10\sin\theta[/tex].
 
  • #3
SpicyPepper said:
Hey people, first time posting. This isn't really homework, but I want to know how to apply inverse trig functions. So, while looking over some problems in my textbook, I came across the following, which I really have no idea how to do.

Homework Statement


A ladder 10ft long leans against a wall. The bottom of the ladder (on the floor) slides away from the wall at 2ft/sec. How fast is the angle between the ladder & the wall changing when the bottom of the ladder is 6ft from the base of the wall?

Homework Equations



The Attempt at a Solution


Since the ladder is 10ft (hypothenuse), the distance from the ladder to wall is 6ft (opposite), then I can figure the wall is effectively 8ft (adjacent).

I assume the angle [tex]\Theta[/tex] can be figured with any inverse trig function.

I know the answer (1/4 rad/s), but that's not really important. I want to be able to understand how to approach this type of problem.

This is a "related rates" problem. The idea is to use the geometry to get an equation in terms of the relevant variables then differentiate with respect to time. That gives the related rates equation in which you substitute the instantaneous values (the 6 and 8).

You have x^2 + y2 = 100. You are given information about x and want information about theta so it would be good to get the y out and the theta into your equation. In your setup

[tex] y = 10 \cos(\theta)[/tex]

So substitute that into get an equation with x and theta. Differentiate it implicitly remembering that everything is a function of t. Then substitute what you know and solve for what you want to know. :-)
 

1. What is an inverse trig function?

An inverse trigonometric function is a mathematical function that undoes the effects of a trigonometric function. It is used to find the angle measure of a triangle when given the ratio of its sides.

2. Why do we need to apply inverse trig functions?

Inverse trig functions are useful in real-world applications such as navigation, engineering, and physics. They help us find missing angles or distances in a triangle, which is essential in solving various problems.

3. How do you apply inverse trig functions?

To apply an inverse trig function, you need to use the inverse button on your calculator or use the inverse function notation, such as sin^-1, cos^-1, or tan^-1. You also need to know the inverse trig identities and how to use them in different scenarios.

4. What are some common inverse trig functions?

The most frequently used inverse trig functions are sin^-1, cos^-1, and tan^-1. Some other common ones include cot^-1, sec^-1, and csc^-1. These functions are the inverse of the corresponding trig functions: sine, cosine, tangent, cotangent, secant, and cosecant.

5. What are the limitations of applying inverse trig functions?

One limitation of applying inverse trig functions is that they are only defined for certain ranges of input values. For example, sin^-1 and cos^-1 are only defined for input values between -1 and 1, while tan^-1 is only defined for input values between -∞ and ∞. Another limitation is that they can only be used to solve for acute angles, not obtuse or reflex angles.

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