Ladder Problem Calculus
Hard Calculus Rate Problem Dealing With a Sliding Ladder?
A 10 foot ladder is sliding down a wall (against which it is leaning) at the rate of 1 in/min. How fast is the foot of the ladder moving away from the wall at the instant when it is two feet from tthe wall?
When you’re dealing with related rate problems, you’re given the rate that one thing changes and asked to find the rate at which something else changes. You do them by first finding a relationship between the two things (not their rates of change) and then taking the time-derivative of both sides.
In your case, you have a 10 foot ladder leaning up against the wall. If you make a sketch, you’ll see that you have a right triangle with a hypotenuse of 10 feet. Let the distance from the base of the wall to the foot of the ladder be x, and let the distance up the wall be y. In that case, you’re given dy/dt and asked for dx/dt, so we need a relationship between x and y. The relationship is obvious – use the Pythagorean Theorem:
x^2 + y^2 = 10^2
Now take d/dt of both sides:
2x dx/dt + 2y dy/dt = d/dt(100)
Since 100 is a constant, the derivative is 0, so
2x dx/dt + 2y dy/dt = 0
x dx/dt + y dy/dt = 0
dx/dt = -(y/x) dy/dt
You know that dy/dt is -1 inch/minute. (Do you see why it’s negative?) You know they want dx/dt when x is 2 feet. If x is 2 feet, then y will be
y = sqrt(100 – x^2) = sqrt(100 – 2^2) = sqrt(96) ~ 9.80 feet
Normally, you’d need to be concerned about units, but since the feet will cancel out anyway, we’ll just ignore it. So:
dx/dt = -(y/x) dy/dt = -(9.8/2) (-1 in/min) = 4.90 in/min
Do you understand the significance of the fact that the answer is positive? Does this make sense, given the conditions of the problem?
Calculus: Related Rates (Ladder Sliding Down a Wall)
