Asked by John
The top of a 13 foot ladder is sliding down a vertical wall at a constant rate of 4 feet per minute. When the top of the ladder is 5 feet from the ground, what is the rate of change of the distance between the bottom of the ladder and the wall
Answers
Answered by
Reiny
In just about any introductory Calculus book this type of question appears as a lead-in to the topic of "rates of change"
Make a diagram, label the height y and the the base length x
then x^2 + y^2 = 13^2
differentiate with respect to t (time)
2x dx/dt + 2y dy/dt = 0
when y = 5 , dy/dx = -4 (negative to show it is getting smaller)
and x^2 + 5^2 = 13^2
x = 12
2x dx/dt = - 2y dy/dt
dx/dt = - (y/x) dy/dt = -5/12 (-4) = 5/3 ft/min
(notice dx/dt is positive, showing the distance to be increasing)
Make a diagram, label the height y and the the base length x
then x^2 + y^2 = 13^2
differentiate with respect to t (time)
2x dx/dt + 2y dy/dt = 0
when y = 5 , dy/dx = -4 (negative to show it is getting smaller)
and x^2 + 5^2 = 13^2
x = 12
2x dx/dt = - 2y dy/dt
dx/dt = - (y/x) dy/dt = -5/12 (-4) = 5/3 ft/min
(notice dx/dt is positive, showing the distance to be increasing)
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