Question

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

Answer 1

The 13 ft ladder is always going to stay the same in this problem and we know it's sliding down a wall at 5 ft/sec

Answer 2

You in calculus?

Answer 3

Yeah, and I know that the 13 ft ladder is consistent and not changing. Just not sure of where to place all the values on the triangle (like what's DX/DT or DY/DT).

Answer 4

First remember the area is \[
A=xy
\]Now remember that \(z\) tells us the relationship between \(x\) and \(y\):\[
z^2=x^2+y^2
\]We already know \(z=13\), so: \[
y^2=13^2-x^2
\]

Answer 5

The area is A = 1/2bh.

Answer 6

We can differentiate with respect to time: \[
2A=xy \implies 2A'=x'y+xy'
\]

Answer 7

Okay, change to \(2A=xy\) then.

Answer 8

sorry I have to think quick

Answer 9

You don't use area you use Pythagorean theorm

Answer 10

We can get \(y'\) using our other equation:\[
y^2=13^2-x^2\implies 2yy' = -2xx' \implies y'= -\frac{xx'}{y}
\]

Answer 11

We can get \(y\) using: \[
y^2=13^2-x^2\implies |y|=y=\sqrt{13^2-x^2}
\]

Answer 12

They already gave us \(x\) and \(x'\), so we have enough information now to get \(y\), \(y'\), and finally \(A'\).

Answer 13

Ummm you don't have to make it this complicated you know x=12 12^2+y^2=13^2

Answer 14

et tu, brute?

Answer 15

And thAn differentiate A=1/2(bh)

Answer 16

Still confused...

Answer 17

Da/dt=1/2(dh/dt(b)+db/dt(h))

Answer 18

Base is db/dt so db/dt=-5

Answer 19

No positive 5

Answer 20

So so far we have da/dt=1/2(dh/dt(12)+5(h))

Answer 21

Did you find dh/dt already?

Answer 22

No, but isn't h 13?

Answer 23

That's the hypotense were looking for the height opposite of the angle which would be side y, and that's not good that would start a whole new problem

Answer 24

Then what's the purpose of 13?

Answer 25

We can find dh/dt quickly though we have to, to find the answer to this problem

Answer 26

ok

Answer 27

So we know 13 stays all the same so we can plug that into the pythgegeon theorm x^2+y^2=z^2 so let's call z=13 so we can plug that in because it will always be the same x^2+y^2=13 now differentiate

Answer 28

so y = 5?

Answer 29

Differentiate

Answer 30

Oh, no. we have to find DY/DT

Answer 31

ok

Answer 32

With respect to time

Answer 33

So you will get?

Answer 34

2x Dx/dt+?

Answer 35

Yep! DY/DT = 2x(DX/DT)

Answer 36

Huh?

Answer 37

2y(DY/DT) = 2x(DX/DT)?

Answer 38

Y^2 move the two out to the front subtract one from power 2y than differentiate with respect to time son
2ydy/dt so you have 2x Dx/dt+ 2y dy/dt =0 now put one of them on the other side

Answer 39

Oh yes sorry one just have to be -

Answer 40

Cause you moved it to the other side

Answer 41

so -2xDX/DT?

Answer 42

Now we can do Pythagorean theorm to find y for the height

Answer 43

That works

Answer 44

isn't y = 5?

Answer 45

Yes now we have all numbers somplug everything in

Answer 46

So Dx/dt = 5 y=5 x=12

Answer 47

so -12?

Answer 48

That's what I got just don't forget units now we need to plug that in for the second part and we now know y=5

Answer 49

so -119/2 ft2/sec?

Answer 50

Thanks a bunch!

Answer 51

That's what I got :)

Answer 52

Did you get it?

Answer 53

yep!

Answer 54

Good work @ducksonquack that was a really long process and you did a good job :)

Answer 55

The triangle lengths give
\[y = \sqrt{13^2-x^2}\]

Writing area in tersm of just the base

\[A = x \sqrt{13^2-x^2}\]

Differentiate w.r.t. time, dA/dt= ...
you have the change in area in terms of just the base x and the rate of change of x now...

Answer 56

Yeah but it gives you the moment in time when the ladder is away from the house

Answer 57

They give you the instant to calculate, when x=12 and dx/dt=5

Answer 58

Right

Answer 59

That is most of these related rate problems from the book. rewrite in terms of single variable and differentiate the equation. Then use the info to calculate the instant in time they want.

Answer 60

I have the same question the teacher gave us the answer key just answers and these were both the answers