Answer:
The pressure is changing at ![\frac{dP}{dt}=3.68](https://tex.z-dn.net/?f=%5Cfrac%7BdP%7D%7Bdt%7D%3D3.68)
Step-by-step explanation:
Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.
We know that the volume is decreasing at the rate of
and we want to find at what rate is the pressure changing.
The equation that model this situation is
![PV^{1.4}=k](https://tex.z-dn.net/?f=PV%5E%7B1.4%7D%3Dk)
Differentiate both sides with respect to time t.
![\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdt%7D%28PV%5E%7B1.4%7D%29%3D%20%5Cfrac%7Bd%7D%7Bdt%7Dk%5C%5C)
The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:
![\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7B%7Bdx%7D%7D%5Cleft%28%20%7Bf%5Cleft%28%20x%20%5Cright%29g%5Cleft%28%20x%20%5Cright%29%7D%20%5Cright%29%20%3D%20f%5Cleft%28%20x%20%5Cright%29%5Cfrac%7Bd%7D%7B%7Bdx%7D%7Dg%5Cleft%28%20x%20%5Cright%29%20%2B%20%5Cfrac%7Bd%7D%7B%7Bdx%7D%7Df%5Cleft%28%20x%20%5Cright%29g%5Cleft%28%20x%20%5Cright%29)
Apply this rule to our expression we get
![V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0](https://tex.z-dn.net/?f=V%5E%7B1.4%7D%5Ccdot%20%5Cfrac%7BdP%7D%7Bdt%7D%2B1.4%5Ccdot%20P%20%5Ccdot%20V%5E%7B0.4%7D%20%5Ccdot%20%5Cfrac%7BdV%7D%7Bdt%7D%3D0)
Solve for ![\frac{dP}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7BdP%7D%7Bdt%7D)
![V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}](https://tex.z-dn.net/?f=V%5E%7B1.4%7D%5Ccdot%20%5Cfrac%7BdP%7D%7Bdt%7D%3D-1.4%5Ccdot%20P%20%5Ccdot%20V%5E%7B0.4%7D%20%5Ccdot%20%5Cfrac%7BdV%7D%7Bdt%7D%5C%5C%5C%5C%5Cfrac%7BdP%7D%7Bdt%7D%3D%5Cfrac%7B-1.4%5Ccdot%20P%20%5Ccdot%20V%5E%7B0.4%7D%20%5Ccdot%20%5Cfrac%7BdV%7D%7Bdt%7D%7D%7BV%5E%7B1.4%7D%7D%20%5C%5C%5C%5C%5Cfrac%7BdP%7D%7Bdt%7D%3D%5Cfrac%7B-1.4%5Ccdot%20P%20%5Ccdot%20%5Cfrac%7BdV%7D%7Bdt%7D%7D%7BV%7D%7D)
when P = 23 kg/cm2, V = 35 cm3, and
this becomes
![\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68](https://tex.z-dn.net/?f=%5Cfrac%7BdP%7D%7Bdt%7D%3D%5Cfrac%7B-1.4%5Ccdot%20P%20%5Ccdot%20%5Cfrac%7BdV%7D%7Bdt%7D%7D%7BV%7D%7D%5C%5C%5C%5C%5Cfrac%7BdP%7D%7Bdt%7D%3D%5Cfrac%7B-1.4%5Ccdot%2023%20%5Ccdot%20-4%7D%7B35%7D%7D%5C%5C%5C%5C%5Cfrac%7BdP%7D%7Bdt%7D%3D3.68)
The pressure is changing at
.
Problem: 2/3= 18/(x+5)
First, I would multiply the (x+5) over
Making it 2/3(x+5)=18
Second, I would Divide by 2/3 (which is the same as multiplying by the reciprocal 3/2)
which shows up like this now (x+5)=18*3/2
third I would subtract the 5 over
which will give you your answer