# Calculus I

You must be logged in to reply to this topic.

• E2P10
• #13249

Ezra Halleck
Participant

3.11: The bottom of a 2 meter ladder leaning against a wall is sliding away from the wall at a constant rate of m/sec. At , the height of the projection of the ladder on the wall is 1 meter. Find the velocity of the top of the ladder at sec.

I suggest that you create two variables that depend on time. Let x represent the position away from the wall of the bottom of the letter and y the position of the top of the latter along the wall. Use the Pythagorean theorem and take the derivative with respect to time implicitly.

#16860

Ezra Halleck
Participant

let x represent position of bottom of ladder (distance to wall) and y represent the position of the top of the ladder (distance to the floor). Note that x^2+y^2=2^2=4. Taking the derivative we get 2x x’ +2y y’ =0. Solving for y’: y’=-x x’/y.
When y=1, x=sqrt(4-1)=sqrt(3). Hence after 1 s, the x=sqrt(3)+.1 and y= sqrt(4- (sqrt(3)+.1)^2)
y’=-(sqrt(3)+.1)(.1)/sqrt(4- (sqrt(3)+.1)^2)

Viewing 2 posts - 1 through 2 (of 2 total)

You must be logged in to reply to this topic.