Question

A 10 ft ladder is being pulled away from a wall at a rate of 3 ft/sec. What is the rate of change in the area beneath the ladder when the ladder is 6 ft from the wall

Answer 1

Answer:

The rate of change in the area beneath the ladder is 5.25 ft²/s

Step-by-step explanation:

Area of triangle is given by;

$A = \frac{1}{2}bh\\\\2A = bh\\\\take \ derivative \ of \ both \ sides \ with \ respect \ to \ "t"\\\\2\frac{dA}{dt} = h\frac{db}{dt} + b\frac{dh}{dt}\\\\divide \ through \ by \ 2\\\\\frac{dA}{dt} = (\frac{h}{2} )\frac{db}{dt} + (\frac{b}{2}) \frac{dh}{dt}$

where;

b is the base of the triangle, given as 6 ft

h is the height of the triangle, determined  by applying Pythagoras theorem.

h² = 10² - 6²

h² = 100 - 36

h² = 64

h = √64

h = 8 ft

Determine the rate of change of the height;

$h^2 + b^2 = 10^2\\\\h^2 + b^2 =100\\\\2h\frac{dh}{dt} + 2b\frac{db}{dt} =0\\\\h\frac{dh}{dt} + b\frac{db}{dt} =0\\\\h \frac{dh}{dt} = -b\frac{db}{dt} \\\\\frac{dh}{dt} = (\frac{-b}{h} )\frac{db}{dt}\\\\\frac{dh}{dt} =(\frac{-6}{8})(3)\\\\\frac{dh}{dt} = -\frac{9}{4} \ ft/s$

Finally, determine the rate of change of area beneath the ladder;

$\frac{dA}{dt} = (\frac{h}{2} )\frac{db}{dt} + (\frac{b}{2}) \frac{dh}{dt}\\\\\frac{dA}{dt} = (\frac{8}{2} )(3) + (\frac{6}{2}) (\frac{-9}{4})\\\\\frac{dA}{dt} = 12 - \frac{27}{4} \\\\\frac{dA}{dt} = \frac{48-27}{4}\\\\\frac{dA}{dt} = \frac{21}{4} \ ft^2/s\\\\\frac{dA}{dt} = 5.25 \ ft^2/s$

Therefore, the rate of change in the area beneath the ladder is 5.25 ft²/s