Question

A 20-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate of 10 feet per second, at what rate is the area of the triangle formed by the wall, the ground, and the ladder changing, in square feet per second, at the instant the bottom of the ladder is 12 feet from the wall?

Answer 1

PLEASSSSSE HELP

Answer 2

the rate of change of area eh

Answer 3

Whats our formula for the area of a rt triangle?

Answer 4

a = 1/2 b*h

Answer 5

good, and when we derive it implicitly with respect to time? we get?

Answer 6

im not sure....is it 1/2 b*h ^-1/2

Answer 7

\[A = \frac{1}{2}bh\]
\[A' = \frac{1}{2}b'h+\frac{1}{2}bh'\]
its just the product rule for derivatives

Answer 8

oh i see

Answer 9

but we can clean it up if we relate h in terms of b since it doesnt give us a rate of change for h does it

Answer 10

so wat is h in terms of b

Answer 11

it might be a trig function, which we might wanna stay away from; OR we could devise a scheme to find h' from b' ...

Answer 12

i got it, pythag

Answer 13

or h=16

Answer 14

\[20^2 = h^2 + b^2\]
\[20^2-b^2 = h^2\]
derived we get
\[-2b\ b' = 2h\ h'\]
\[\frac{-2b\ b'}{2h} = h'\]
\[\frac{-b\ b'}{h} = h'\]

Answer 15

this is doable

Answer 16

\[A' = \frac{1}{2}b'h+\frac{1}{2}bh'\]

\[A' = \frac{1}{2}b'h+\frac{1}{2}b(\frac{-b\ b'}{h})\]

messy, but doable; now we are told b' = 10, and b = 12

the pythag itself will give us our h at this moment in time and then its just plug and play

Answer 17

wat do i plug for h

Answer 18

\[20^2 = 12^2 + h^2\]
\[400-144 = h^2\]
\[\sqrt{256} = h\]
\[16 = h\] right?

Answer 19

ys

Answer 20

then all together we get:

\[A' = \frac{1}{2}b'h+\frac{1}{2}b(\frac{-b\ b'}{h})\]
\[A' = \frac{1}{2}10(16)+\frac{1}{2}12(\frac{-12(10)}{16})\]
\[A' = 80+6(\frac{-12(10)}{16})\]
\[A' = 80+(\frac{-2(10)}{16})\]
\[A' = 80-\frac{10}{8}\]

Answer 21

the aswer choices are
a) -70, b)35/2
c) 70 d) -35
e) 35

Answer 22

wen u solve for wat u did isnt teh asnwer 78.75

Answer 23

lets look over our math to be sure

Answer 24

i get 35 when I do the calculations right

Answer 25

thanks :)

Answer 26

wat didu do wrng in the begining

Answer 27

\[A' = \frac{1}{2}10(16)+\frac{1}{2}12(\frac{-12(10)}{16})\]
\[A' = 10(8)+12(\frac{-12(5)}{16})\]
\[A' = 80+3(-3(5))\]
\[A' = 80-45=35\]

Answer 28

the formulas were good, I just forgot how to multiply and add lol