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Math Help - Calc. Word Problems

  1. #1
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    Calc. Word Problems

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

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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    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
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