Applying inverse trig function

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To apply inverse trig functions in related rates problems, start by establishing the relationship between the variables using geometry. In this case, with a 10ft ladder leaning against a wall, the bottom slides away at 2ft/sec, and when 6ft from the wall, the height can be calculated as 8ft. The equation x² + y² = 100 can be differentiated with respect to time to find the rate of change of the angle θ. Substitute y = 10cos(θ) into the equation to express it in terms of x and θ, then differentiate implicitly. This method allows for solving for the rate at which the angle is changing effectively.
SpicyPepper
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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 \Theta 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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Since the triangle stays in contact with the wall, you can say that x^{2}+y^{2}=10^{2} at all times, and x=10\cos\theta, y=10\sin\theta.
 
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 \Theta 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

y = 10 \cos(\theta)

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. :-)
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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