Question

A ladder 25 feet long is leaning against the wall of a house. The base of the ladder is pulled away from the wall at a rate of 2 feet per second.

Answer 1

Is this all the info?

Answer 2

You may want to draw a picture.

Answer 3

Part A. How fast is the top of the ladder moving down the wall when the base of the ladder is 7 feet from the way?

Answer 4

However, I am guessing you use : \[
z^2=x^2+y^2
\]

Answer 5

I have 2 other parts.
I have solve part A. \[\frac{ db }{ dt}=-12.4375 \]

Answer 6

feet per second

Answer 7

Part B:
Consider the triangle formed by the side of the house, the ladder, and the ground. Find the rate at which the area of the triangle is changing when the base of the ladder is 7 feet from the wall.

Answer 8

I used \[A=\frac{ 1 }{ 2}bh\]

Answer 9

\[\frac{ dA }{ dt }=\frac{ 1 }{ 2 }h \frac{ db }{ dt}+\frac{ 1 }{ 2 }b\frac{ dh }{ dt}\]

Answer 10

\[\frac{ dA }{ dt }=-142.25\]

Answer 11

feet^2/second
not sure this solution is correct

Answer 12

You have b, h, b', h', right? so what is the issue?

Answer 13

I am not sure I did this one correctly nor the last part which involves the angle changing

Answer 14

Which part are you unsure about?

Answer 15

the last part the solution and the units I tend to get the units incorrect

Answer 16

area/time is feet^2/sec

Answer 17

thanks did I get the solution correct as well?

Answer 18

I know I needed to do part A first because I needed that solution to figure out Part B

Answer 19

Part C I am really stuck on
Find the rate at which the angle formed by the ladder and the wall of the house is changing when the base of the ladder is 9 feet from the wall

Answer 20

Hmm, what did you get for h, b, h'?

Answer 21

h=24 b=7 h'=2 because that was given and b'=-12.4375 which I solved for in part A

Answer 22

The base is the opposite side to the angle formed by the wall and latter. \[
25\sin(\theta) = b
\]

Answer 23

ok so now we are on to the last part, correct. I am sorry I was still answering part b but that is fine, I can move on to part C

Answer 24

is this the angle I am looking for?

Answer 25

I would think so, since it is formed between wall and ladder.

Answer 26

Can I use sine , cosine or tangent?

Answer 27

\[\sin \theta=\frac{ x }{ 25 }\]

Answer 28

Yes

Answer 29

I was going to take the derivative of it

Answer 30

\[\cos \theta \frac{ d \theta }{ dt}=\frac{ 1 }{ 25}\frac{ dx }{ dt}\]

Answer 31

\[
\cos\theta = \frac y{25}
\]

Answer 32

\[\cos \theta=\frac{ \sqrt{544} }{ 25}\]

Answer 33

I can sub for cosine theta, but I don't know dx/dt unless I use the same procedure for part A

Answer 34

Not to be the bearer of bad news you need to recalc your answer for part a.

Answer 35

Is part A incorrect?

Answer 36

I never saw your work for A, and I didn't check it either. What did you do?

Answer 37

625 = x^2 +y^2

0 = 2xx' + 2yy'

y' = -xx'/y = -14/24

Answer 38

yes d/dx of 625 is 0

Answer 39

d/dt of constant = 0

Answer 40

that will change part B as well

Answer 41

Thanks part A is -7/12 ft/sec

Answer 42

Yep

Answer 43

Your process is right just wrong numbers.

Answer 44

so is part b 0?

Answer 45

\[\frac{ dA }{ dt}=\frac{ 1 }{ 2 }h \frac{ db }{ dt }+\frac{ 1 }{ 2 }b \frac{ dh}{ dt }\]

Answer 46

2a' = yx' + xy'

a' = (24(2) + 7(-14/24))/2 = 527/49 ft^2/sec

Answer 47

\[\frac{ dA }{ dt }=\frac{ 1 }{ 2 }(24)\frac{ -7 }{ 12 }+\frac{ 1 }{ 2 }(7)(2) \]

Answer 48

opps that should be 24 in the denom :P

Answer 49

instead of 49

Answer 50

24 times 2?

Answer 51

sorry I have to relabel my triangle using h and b

Answer 52

I believe you are correct

Answer 53

Then for c \[\sin(\theta) = x/25\]

\[\cos(\theta)\theta' = x'/25\]

Answer 54

\[\theta' = 2/(25\cos(.284)) \]

Answer 55

ok you lost me with cos how did you get .284? I understood all of the rest. I am using dt notation but I can follow along with your notation

Answer 56

\[\sin(\theta) = (7/25)\]

Answer 57

\[\theta = \sin^-1(7/25) =.284\]

Answer 58

does it matter on part c that it was changed to 9 feet from the wall? Let me rewrite part c

Answer 59

Part C. Find the rate at which the angle formed by the ladder and the wall of the house is changing when the base of the ladder is 9 feet from the wall

Answer 60

I know the ladder is 25

Answer 61

Oh yeah it would change everything I did not see that.

Answer 62

ok thanks I think I got it from here. I will post my final solution so you can double check my answer since I tend to mess these up

Answer 63

Ok no problem :D

Answer 64

Actually you do not need to recalc it would be the same just switch out 7 for 9 when determining the angle then use the equation as x' would remain constant. Sorry I was thinking about how it would affect y :P

Answer 65

is dx/dt the given rate of 2 feet per second?

Answer 66

that is ok, I have been doing these incorrect because I am making small mistakes

Answer 67

ok I got .0857492926

Answer 68

for d(theta)/dt

Answer 69

rad per second

Answer 70

Looks good :)

Answer 71

thank you so much I really appreciate this

Answer 72

Yep :)

Answer 73

A solution using Mathematica is attached.