Rate of Change of theta in ladder to wall.

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SUMMARY

The discussion focuses on calculating the rate of change of the angle (theta) between a 10 ft ladder and the ground as the base of the ladder slides away from a wall at a rate of 1.3 ft/s. When the bottom of the ladder is 6 ft from the wall, the angle's rate of change can be determined using trigonometric relationships. The correct approach involves using the equation a = b cos(theta) and the tangent function to express theta in terms of the distances involved. The solution requires applying differentiation to find the rate of change of theta with respect to time.

PREREQUISITES
  • Understanding of basic trigonometry, specifically sine, cosine, and tangent functions.
  • Familiarity with the Pythagorean theorem and its application in right triangles.
  • Knowledge of implicit differentiation and its use in related rates problems.
  • Ability to apply calculus concepts to real-world problems involving motion and angles.
NEXT STEPS
  • Study the application of implicit differentiation in related rates problems.
  • Learn how to derive angles using trigonometric identities and relationships.
  • Practice solving similar problems involving ladders and walls to reinforce understanding.
  • Explore the concept of angular velocity and its applications in physics.
USEFUL FOR

Students studying calculus, particularly those focusing on related rates problems, as well as educators teaching trigonometry and its applications in physics and engineering contexts.

ryandamartini
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Homework Statement


A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.3 ft/s, how fast is the angle between the ladder and the ground changing when the bottom of the ladder is 6 ft from the wall? Evaluate your answer numerically.


Homework Equations



a^2+b^2=c^2 ?
Theta(dot) is change of time respect to theta.

The Attempt at a Solution



In class he went over finding the rate of theta, getting the equation Theta(dot)=X/(10cos(45) I tried plugging in 6 for X, I just really don't know what to do here.
 
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I would suggest that instead of using a2+ b2= c2, which does not involve an angle, you use a= bcos(\theta) where a is the "hypotenuse" (the length of the ladder) and a is the "near side" (the distance from the foot of the ladder to the wall).
 
theta=tan(y/x) right
then what the formula to find the change in theta at any given point?
 

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