Question

A ladder 25 feet long is leaning against a house. The base of the ladder is pulled away from the house wall at a rate of 2feet per second. How fast is the top of the ladder moving down the wall when the base of the ladder is 7feet from the wall? (Calculus differentiation)

Answer 1

a^2+b^2=c^2
a=7; da/dt=2 find b and db/dt
b^2=(25)^2-(7)^2
b=24
a(da/dt)+b(db/dt)=0
7(2)+24(db/dt)=0
db/dt=-14/24
db/dt=-7/12
so decreasing at a speed of 7/12

Answer 2

in feet

Answer 3

why do we use this equation?
a(da/dt)+b(db/dt)=0

Answer 4

u have \[a^2+b^2=c^2\]
and the ladder is moving at a rate so we kno that we have to take the derivative implicitly with respect to time. now we need to figure out the variables.

when a ladder is leaned against the wall u can think of the ladder as the hypotenuse of the triangle. also the ladder isnt changing so not only is it the hypotenuse we kno it is a constant so c must b a constant. so when we take the derivative we kno it will b 0
a and b however are changing. as the ladder falls the side of the building that makes a triangle with the ladder is getting smaller. we are actually trying to find how fast it is decreasing right now.

since we know how fast the sidealong the ground is increasing and we have the Pythagorean relationship when u take the derivative implicitly the relationship between their speeds has to b related. In this case its related by that formula

Answer 5

in less words:
its the formula that shows the relationship between the speeds of the sides when the hypotenuse is a constant

Answer 6

thanks! i get it now :D