Calculus question

Question

A 7m ladder is leaning against the wall. When the top of the ladder is slipping down at a rate of 2m/s, how fast is the bottom moving when it is 4m away from the wall?

 

HELP!?

Answer 1

Here is what we know:

 

a² + b² = c²

 

where:

c = length of ladder

a = vertical length

b = horizontal length

 

c = 7

b = 4

a = √(7² - 4²) = √(49 - 16) = √33

db/dt = ?

da/dt = -2

dc/dt = 0  ---> length of ladder is constant

 

 

Differentiate both sides of the Pythagorean theorem equation.

 

d/dt(a² + b²) = dc/dt

 

2a(da/dt) + 2b(db/dt) = 0

 

2√(33)(-2) + 2(4)(db/dt) = 0

 

-4√(33) + 8(db/dt) = 0

 

 

Now just solve for db/dt.  This is the rate at which the bottom is moving.

Answer 2

 

 

First, assuming the wall is perpendicular to the ground, you can use the Pythagoras equation:

 

C²=A²+B² where C=hypotenuse A= height and B=length from the wall.

 

Lets get the length of A when B=4m:

 

7²=A²+4² --> 49=A²+16--> A²=33 --> A=(33)^1/2

 

 Isolate the variable you want, which is the bottom speed moving away from the wall (dB), by first isolating the B variable itself (also remember that C is a constant unless you have a stretchy ladder):

 

B= (C² - A² ) ^1/2 --> B= (7² - A² )^1/2

 

Now you can take the derivative of the entire equation (using chain rule):

 

dB=1/2(49-A²)^-1/2 x (-2A )dA --> dB=-A(49-A²)^-1/2dA

 

Now that you have the dB (how fast it is moving away from the wall) and dA is already given to you as 2m/s down so dA=-2m/s

 

Thus now when you input in all in since A=(33)^1/2 and dA=-2:

 

dB=-(33)^1/2(49-(33^1/2))^-1/2(-2). Solve for dB and you are golden.

 

 

Answer 3

You have a right triangle with a constant hypotenuse of 7m, a height h shrinking at a rate of 2 m/s, and a base x that is growing.

 

dh/dt = -2 m/s

 

x = √(7² - h²) = √(49 - h²)

 

dx/dt = (1/2)(49 - h²)^-1/2 (-2h dh/dt)

 

When x = 4,

 

h = √(49 - 4²) = √33 ≅ 5.74 m

 

Then when x = 4,

 

dx/dt = (1/2)(49 - 33)^-1/2(-2)(5.74)(-2 m/s) = 2.87 m/s