Related Rate Problem & defferential

[packerman]

I am working on a couple of problems for my calc class, and I was wondering if the answer I got for the rate of horizontal change in problem 1 is correct and if the answers I got in problem 2 are correct.

========Problem 1=========

A winch at the top of a 12-meter building pulls a pipe of the same length to a vertical position as shown (picture of building with pole at its base, forming a triangle of 12-meter building, 12-meter pole, and side "s" made of rope). The winch pulls in rope at a rate of -0.2 meters per second. Find the rate of vertical change and rate of horizontal change at the end of the pipe when y=6.


Rate of Horizontal Change:

x²+y²=12²

(This step really confuses me) :arrow: s²=y²+(12 + x)² = 288+24x

2sds=24dx

dx/ds=2s/24=
s/12 =
sqrt(288+24x)/12

ds/dt=-0.2

dx/ds*ds/dt = dx/dt = sqrt(288+24x)/12)*(-0.2) =

sqrt(288+24x)/60

When y=6 ; x= sqrt(108)

dx/dt = sqrt(288+24*sqrt(108))/60 = .38637 m/s


========Problem 2=========
The profit P for a company is given by P(x) = (500x - x^2) - (1/2x^2 -77x +3000) where x is the number of units produced.

(a) Use differentials to approximate the change in profit as production changes from x=115 to x=120 units.

(b) What is the approximate precent change?


(a) P'(x) = 577 - 3x; P(115) = 43517.5; P(120) = 44640

dy = P'(x) dx = (577 - 3x)dx

when x = 115 and dx=5

dy=(577-3(115))-5=1160

When the number of units produced increases from 115 to 120 there will be a $1160 increase in profit.


b. (really not sure about this one....)


P(120) / P(115) = 1.02579

There will be a 1.03% gain in profit when the unit production changes from 115 to 120

Any help is appreciated!

 

[soroban]

Hello, packerman!

I hope I interpreted this problem correctly . . .

A winch at the top of a 12-meter building pulls a pipe of the same length to a vertical position.
The winch pulls in rope at a rate of -0.2 meters per second.
Find the rate of vertical change and rate of horizontal change at the end of the pipe when y = 6.

                    * A
                    |
                    |12-y
                    |
                    * Q
                  / |
             12 /   |
              /     |y
            /       |
          * - - - - * 
          P    x    B

The building is \(\displaystyle AB\,=\,12\)

The 12-m ladder reaches from \(\displaystyle P\) on the ground to \(\displaystyle Q\) on the building.

Let \(\displaystyle x\,=\,PB.\)

Let \(\displaystyle y\,=\,QB\) . . . then \(\displaystyle 12\,-\,y\:=\:AQ\)


From Pythagorus, we have: \(\displaystyle \,x^2\,+\,y^2\:=\:12^2\)

Differentiate with respect to time: \(\displaystyle \:2x\left(\frac{dx}{dt}\right)\,+\,2y\left(\frac{dy}{dt}\right)\:=\:0\)

\(\displaystyle \;\;\)and we have: \(\displaystyle \L\,\frac{dx}{dt}\;=\;-\frac{y}{x}\left(\frac{dy}{dt}\right)\)


When \(\displaystyle y\,=\,6\), we have: \(\displaystyle \,x^2\,+6^2\:=\:12^2\;\;\Rightarrow\;\;x\,=\,\sqrt{108}\,=\,6\sqrt{3}\)

Since the rope is pulled at -0.2 m/sec, \(\displaystyle y\) is increasing at the same rate.
\(\displaystyle \;\;\)That is: \(\displaystyle \,\frac{dy}{dt}\,=\,+0.2\,=\,\frac{1}{5}\) m/sec.


Therefore: \(\displaystyle \L\:\frac{dx}{dt}\;=\;-\frac{6}{6\sqrt{3}}\left(\frac{1}{5}\right)\;=\;-\frac{\sqrt{3}}{15}\) m/sec \(\displaystyle \;\)(rate of horizontal change).

And we already know that the rate of vertical change is: \(\displaystyle \,\frac{dy}{dt}\,=\,0.2\) m/sec.

 

[packerman]

The picture:

 

[galactus]

By the triangles, we can find x when y=6.

\(\displaystyle \L\\x=\sqrt{144-36}=\sqrt{108}=6\sqrt{3}\)

Therefore, \(\displaystyle s=\sqrt{(6\sqrt{3})^{2}+6^{2}}=12\)

\(\displaystyle \L\\x^{2}+36=s^{2}\)

\(\displaystyle \L\\2x\frac{dx}{dt}=2s\frac{ds}{dt}\)

\(\displaystyle \L\\(6\sqrt{3})\frac{dx}{dt}=(12)(-.02)\)

\(\displaystyle \L\\\frac{dx}{dt}=\frac{\frac{-12}{5}}{3\sqrt{6}}=\frac{-2\sqrt{3}}{15}=\)-.231 m\sec

Please check my work. It's easy to err in related rates problems.