Question

related rates problem

Answer 1

A 13ft ladder is leaning against a house when its base starts to slide away. By the time the base is 12ft from the house the base is moving at the rate of 5ft/sec b) at what rate is the area of the triangle formed by the ladder, wall, and ground changing at that moment

Answer 2

In part a I found da/dt (height change) to be 12 ft/sec

Answer 3

I got negative -12 when I did the work in part a though

Answer 4

I think that the book said just 12 because they were just giving the magnitude and ignoring direction

Answer 5

It should be -12 to be exact, but if they're looking at the rate of decrease, then yes, 12 is the right answer.

Answer 6

ok so how do I do b because they say the answer is -119/2 ft^2/sec

Answer 7

I used A1/2bh

Answer 8

did implicit differentiation using the product rule and got\[\frac{ dA }{ dt }=\frac{ 1 }{ 2 }a \frac{ da }{ dt }+b \frac{ db }{ dt }\]

Answer 9

but that ended up giving me -120/2

Answer 10

we know a is 5 from pythagorean

Answer 11

\[A = \frac{ab}{2}\]
\[2\frac{dA}{dt} = a \frac{db}{dt} + b \frac{da}{dt}\]
\[\frac{dA}{dt} = \frac{5(5) + 12(12)}{2} = \frac{169}{2}\]
I got a totally different answer....

Answer 12

it still isnt right according to the book
the book says it's -119/2

Answer 13

I know... I'm calling in a third party to check.

Answer 14

Wait, my bad, \[\frac{dA}{dt} = \frac{5(5) + 12(-12)}{2} = -119/2\]

Answer 15

You made a mistake in the product rule formula. It should be \[(uv)' = uv' + vu'\]

Answer 16

I see it now too you're right

Answer 17

could you check me on part c now

Answer 18

it's asking for the rate of change of theta