Calc. Word Problems

ulion · #1 $Old$ September 7th, 2008, 07:43 AM

heya, I absolutely horrible with Calculus Word Problems and was wonder if anyone could help me with these.

Chris L T521 · #2 $Old$ September 7th, 2008, 11:02 AM

Quote:

Originally Posted by ulion $View Post$

heya, I absolutely horrible with Calculus Word Problems and was wonder if anyone could help me with these.

#148

We are told that $\frac{\,dV}{\,dt}$ is inversely proportional to the square of $t+1$

This tells us that $\frac{\,dV}{\,dt}=\frac{k}{(t+1)^2}$ , where $k$ 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.

$\frac{\,dV}{\,dt}=\frac{k}{(t+1)^2}\implies\,dV=k\frac{\,dt}{(t+1)^2}$

Thus, we see that $V=-\frac{k}{t+1}+C$

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 $V(0)=500000$

Applying this condition to the equation we have for V, we see that $500000=-k+C$

Now let us look at the second condition: "Its value decreased by $100,000 in the first year"

This is saying that $V(1)=400000$

Applying this condition to the equation we have for V, we see that $400000=-\frac{k}{2}+C$

We have to solve this system for $k$ and $C$ :

$\left\{\begin{array}{rcr}-k+C&=&500000\\-\frac{1}{2}k+C&=&400000\end{array}\right.$

I leave it for you to verify that $k=-200000$ and $C=300000$

Thus, our equation for V is $V(t)=\frac{200000}{t+1}+300000$

Now all you have to do is find $V(4)$

$V(4)=\frac{200000}{(4)+1}+300000=\dots$

# 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 $\frac{\,dS}{\,dt}=\frac{k}{t}$ , where $k$ is the constant of proportionality.

Using separation of variables, we see that $\,dS=\frac{k}{t}\,dt\implies S=k\ln|t|+C$

We are given two conditions: $S(2)=200$ , and $S(4)=300$ .

Use a similar process to # 148 to get an equation for $S(t)$

# 68

Set up the integral:

$\int_0^{\ln\sqrt{3}}\frac{e^x}{1+e^{2x}}\,dx$

Make the substitution $u=e^x$ .

I leave it for you to verify that:

$\int_0^{\ln\sqrt{3}}\frac{e^x}{1+e^{2x}}\,dx=\int_1^{\sqrt{3}}\frac{\,du}{1+u^2}$

Then evaluate the integral.

I hope this makes sense!

--Chris