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Calculus/Related Rates

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Expert: - 3/22/2009


Question

Diagram of Ladder
A ladder 15 feet long is leaning against a building so that end X is on level ground and end Y is on the wall as shown in the figure. The bottom of the ladder (X) is moved away from the building at a constant rate of ½ foot per second.
a) Find the rate in feet per second at which the length OY is changing when X is 9 feet from the building.

b) Find the rate of change in square feet per second of the area of triangle XOY when X is 9 feet from the building.

So I know I have to use the pythagorean theorem so I statred out with √(225-x(t)^2). Then I have truble finding its deriv. And with (b) do I use the general area equation A(t)=[x(t)*y(t)]2 first with that one?

Answer
Questioner:   Jeanne
Country:  United States
Category:  Calculus
Private:  No
 
Subject:  Related Rates/Change of rate of a 15ft ladder
Question:  

A ladder 15 feet long is leaning against a building so that end X is on level ground and end Y is on the wall as shown in the figure. The
bottom of the ladder (X) is moved away from the building at a constant rate of ½ foot per second.

a) Find the rate in feet per second at which the length OY is changing when X is 9 feet from the building.

b) Find the rate of change in square feet per second of the area of
triangle XOY when X is 9 feet from the building.

So I know I have to use the pythagorean theorem so I statred out with √(225-x(t)^2). Then I have truble finding its deriv.

And with (b) do I use the general area equation A(t)=[x(t)*y(t)]2
first with that one?
..................................
Nice diagram.  No curtains on the windows?

Sorry.

Let  y = length OY
    x = length OX
    A = area of the triangle.

Rates:

dy/dt TO BE FOUND.
dx/dt = 0.5
dA/dt TO BE FOUND.

Relations:

x^2 + y^2 = 15

Diff:  << always use implicit differentiation.  Then you don't have 'truble'.

2x dx/dt + 2 y dy/dt = 0

x dx/dt +  y dy/dt = 0

Values:

dx/dt = 0.5,  x = 9, and, using the P.T.,  y = 12.

(9)(0.5) +  (12) dy/dt = 0

(3)(0.5) +  (4) dy/dt = 0

dy/dt = 1.5/4 = 0.375

.................................

Relations:

A = xy/2, area formula for a triangle.

Diff:

dA/dt = 1/2(x dy/dt + y dx/dt)  << product rule.

Values: (from above)


dx/dt = 0.5,  x = 9, y = 12, dy/dt = 0.375

You can handle the rest.

Paul Klarreich

Expertise

All topics in first-year calculus including infinite series, max-min and related rate problems. Also trigonometry and complex numbers, theory of equations, exponential and logarithmic functions. I can also try (but not guarantee) to answer questions on Analysis -- sequences, limits, continuity.

Experience

I taught all mathematics subjects from elementary algebra to differential equations at a two-year college in New York City for 25 years.

Education/Credentials
(See above.)

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