heya, I absolutely horrible with Calculus Word Problems and was wonder if anyone could help me with these.
#148
We are told that $\displaystyle \frac{\,dV}{\,dt}$ is inversely proportional to the square of $\displaystyle t+1$
This tells us that $\displaystyle \frac{\,dV}{\,dt}=\frac{k}{(t+1)^2}$, where $\displaystyle k$ is a constant of proportionality.
You need to solve this differential equation, and the best way to do so would be with separation of variables.
$\displaystyle \frac{\,dV}{\,dt}=\frac{k}{(t+1)^2}\implies\,dV=k\ frac{\,dt}{(t+1)^2}$
Thus, we see that $\displaystyle V=-\frac{k}{t+1}+C$
Now, this is where two conditions come into play:
The first condition: "The initial value of the machine was $500,000"
This is saying that $\displaystyle V(0)=500000$
Applying this condition to the equation we have for V, we see that $\displaystyle 500000=-k+C$
Now let us look at the second condition: "Its value decreased by $100,000 in the first year"
This is saying that $\displaystyle V(1)=400000$
Applying this condition to the equation we have for V, we see that $\displaystyle 400000=-\frac{k}{2}+C$
We have to solve this system for $\displaystyle k$ and $\displaystyle C$:
$\displaystyle \left\{\begin{array}{rcr}-k+C&=&500000\\-\frac{1}{2}k+C&=&400000\end{array}\right.$
I leave it for you to verify that $\displaystyle k=-200000$ and $\displaystyle C=300000$
Thus, our equation for V is $\displaystyle V(t)=\frac{200000}{t+1}+300000$
Now all you have to do is find $\displaystyle V(4)$
$\displaystyle V(4)=\frac{200000}{(4)+1}+300000=\dots$
# 96
This question is similar to the first one here.
Following the same idea, we see that the equation modeling the number of sales per week is $\displaystyle \frac{\,dS}{\,dt}=\frac{k}{t}$, where $\displaystyle k$ is the constant of proportionality.
Using separation of variables, we see that $\displaystyle \,dS=\frac{k}{t}\,dt\implies S=k\ln|t|+C$
We are given two conditions: $\displaystyle S(2)=200$, and $\displaystyle S(4)=300$.
Use a similar process to # 148 to get an equation for $\displaystyle S(t)$
# 68
Set up the integral:
$\displaystyle \int_0^{\ln\sqrt{3}}\frac{e^x}{1+e^{2x}}\,dx$
Make the substitution $\displaystyle u=e^x$.
I leave it for you to verify that:
$\displaystyle \int_0^{\ln\sqrt{3}}\frac{e^x}{1+e^{2x}}\,dx=\int_ 1^{\sqrt{3}}\frac{\,du}{1+u^2}$
Then evaluate the integral.
I hope this makes sense!
--Chris