Ladder Related Rates

Question

A 30 ft ladder is leaning against the side of a building. You pull the base of the ladder away from the wall of the building at a rate of 9 in/sec.

At what rate is the height of the ladder above the ground changing when the base of the ladder is 10 feet from the wall of the building?

Round to the nearest thousandth when necessary.

Answer 1

h = height of ladder at time t and x = distance from the base of the wall to the ladder's base at time t.

Given: length of ladder = 30 ft and dx/dt = 9 in/sec = (3/4) ft/sec

Find: dh/dt when x = 10 ft

By the Pythagorean Theorem, x² + h² = 900

2x(dx/dt) + 2h(dh/dt) = 0

15 + 2h(dh/dt) = 0

dh/dt = -15 / (2h)

When x = 10, 100 + h² = 900. So, h = √800 = 20√2.

Therefore, dh/dt = -15 / (40√2) = -3 / (8√2) = -3√2 / 16 ≈ -0.265 ft/sec

Answer 2

construct a right triangle

use pythagorean theorem

h^2 = x^2+y^2

take derivatives with respect to time

2hh'= 2xx' +2yy'

divide by 2

hh'= xx'+ yy'

plug in the values

h=30, h'=0, x=10, y= sqr(900-100)=sqr800=20sqr2, x'=9/12=3/4

(30)0= 10(3/4)+(20sqr2)y'

y' = -(15/2)/20sqr2 = -1.5/4sqr2=-0.375/sqr2

y' =-0.1875sqr2

= about- .1875(1.414)

= -.265125

= about -.265 feet per second

= about -3.1815

= about -3.182 inches per second