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\begin{document}

\noindent
{\bf Math 105} \hfil {\bf Answers to test 2}

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\noindent {\bf Part I.  Multiple Choice}

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1. Which of the following expressions is equivalent to $e^{\ln t + \ln (1/t)}$?
\quad {\bf Ans.} 1.  

\smallskip
Because $\ln t + \ln (1/t) = \ln\left(t\cdot{1\over t}\right) = \ln 1 = 0$.

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2. Given a function $y=f(x)$, we can calculate the derivative by using the formula
$$f'(x) = \lim_{h \to 0}{f(x+h)-f(x)\over h} \hbox{\ \  ({\bf Ans.})}$$

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3. If you deposit \$500 in an account paying $4\%$ interest compounded continuously, how much will be in the account
after 20 years? \ {\bf Ans.} $500e^{0.8}$.  From the formula for continuously compounded interest,
$y = Ae^{rt}$, with $A=500$, 
$r= 0.04$ and $t=20$

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4. Which of the following limits would you obtain when using the definition of the derivative to compute
$f'(x)$ where $f(x) = x^2-x$? {\bf Ans.} $\displaystyle \lim_{h \to 0} {2xh+h^2-h \over h}$.  Apply the basic definition (as
given in question 2 above) and simplify.

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5.  What is  the derivative $f'(x)$ of the function $f(x)=3x^4-{4\over{\sqrt x}}\ $?
{\bf Ans.} $12x^3 + 2x^{-3/2}$.  Rewrite $f(x) = 3x^4 - 4x^{-1/2}$ and use the power rule with the addition and coefficient
rules.

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6. A company finds that if it produces and sells  $x$ boxes of assorted chocolates per week, its profit (in dollars) is
$-1000+15x +0.02 x^2$. What is its marginal profit (in dollars) at a production level of $100$ boxes
per week? {\bf Ans.} \$19.00.  The marginal profit function is $P'(x) = 15 + 0.04x$, so $P'(100) = 19$.

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7. If $y=e^{2t}-2e^t$, which of the following functions is equal to ${dy\over dt}\ $?
{\bf Ans.} $2e^{2t} - 2e^t$. Use the formula ${d\over{dt}}\,e^{ct} = ce^{ct}$ with the addition and coefficient formulas.

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8. If $g(x)=3x^4-2x^3+5x-\ln x$, which of the following functions is equal to $g''(x)$ ?

{\bf Ans.} $\displaystyle 36x^2 -12x + {1\over{x^2}}$, because $\displaystyle g'(x) = 12x^3 - 6x^2 + 5 - {1\over x}$.

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9. What is the   point on the curve $y=\ln x$ where the tangent line is parallel to the line $3y=2x+1$?
{\bf Ans.} $ ({3\over 2}, \ln {3\over2})$. Solve the equation $ {1\over x} = {2\over3}$.  

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10. With an inflation rate of $2\%$ per annum, prices are described by the formula $P=P_0(1.02)^t$ where $P$ is the price in
dollars after $t$ years and $P_0$ is the price when $t=0$. How fast are prices rising in dollars per year when $t=3$?
{\bf Ans.} $P_0 (1.02)^3\ln (1.02) $.  Use the formula ${d\over{dt}}\,b^t = b^t\ln b$ with $b = 1.02$, then plug in $t=3$.

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{\bf Part II.  Partial Credit}

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11. Find all  values of $x$ which satisfy the equation $4+\ln (x^3) = 10$.
{\bf Ans.} $x=e^2$ is the only solution.  Rewrite as $\ln(x^3) = 6$, so that $x^3 = e^6$, and
$x = \left(e^6\right)^{1/3} = e^2$.

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12. You are considering investing some money with Investments 'R' Us. When you do some
research, you find that they plan to invest your money in bonds paying 7\% interest compounded
continuously. How long will it take Investments 'R' Us to triple your investment?
{\bf Ans.} $(\ln3)/0.07 \approx 15.7$ years.  With $A$ being the initial
investment and $t$ the time in years, you have to solve the equation
$3A = Ae^{0.07t}$.

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13. Tritium, a synthetic form of hydrogen, satisfies the exponential decay law, $y=Ae^{-kt}$,
where $t$ is the time (in years) and $y$ is the amount (in grams) at time $t$. If an
absent-minded scientist receives a shipment of 200 grams of tritium and finds that only 50
grams of it are left when he remembers the sample 25 years later, find a function that gives
the amount of tritium remaining in the sample $t$ years after the scientist received the
shipment.

{\bf Ans.} $200e^{-t(\ln4)/25}$ or $200e^{t\left(\ln(1/4)\right)/25}$ or (approximate)
$200e^{-0.05545t}$. 

First solve the equation $50 = 200e^{-25k}$ for $k$, then substitute that
value of $k$ into the decay equation, also setting $A = 200$.

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14. Find the equation of the tangent line to the curve $y=2x^3-2x$ at the point $(1,0)$.

{\bf Ans.} $y = 4x - 4$.  First compute the derivative, $y' = 6x^2 - 2$, then plug in $x=1$ to find
the slope at $(1,0)$.  Then write the equation of the line through $(1,0)$ with that slope.

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15. A toy rocket launched straight up from the top of a building  has a height $s$ in feet above
ground given by the formula $s=96 +80t-16t^2$, where $t$ denotes the time in seconds after
launch. \ 
 (a) When is the velocity of the rocket zero?
{\bf Ans.} At $t= 5/2$ sec.  You have to solve the equation $dy/dt = 0$.
\ 
(b)  What is the  velocity of the rocket when it strikes the ground? 
{\bf Ans.} $-112$ ft. per sec.  First solve the equation $96 +80t-16t^2 = 0$ to find when the
rocket strikes the ground.  The only positive solution is $t=6$.  Then plug that into the
derivative $y' = 80 - 32t$.

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16. Let $P(x)$ be the profit made by a company selling $x$ midsize cars per year. Suppose that
$P(100)=60,000$ and $P'(100)=1,000$. Estimate the profit from the sale of $102$ cars per year.
{\bf Ans.} \$62,000.  Use the linear approximation formula 
$P(102) \approx P(100) + P'(100)\cdot2$.

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17. Calculate the derivative of the function $y=\ln{\root 3 \of x}$.
{\bf Ans.} $ y' = {1\over{3x}}$.  Rewrite $ y = {1\over3}\ln x$, so that
$ y' = {1\over3}\,{d\over{dx}}\,\ln x = {1\over{3x}}$

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18. The distance $s$ in feet travelled by a car during the first ten seconds after it starts from
rest is given by the formula $s={5\over 2}t^3-{4\over 7}t^{7/2}$, where $t$ is the time in
seconds. \ (a)  What is the average velocity of the car in the first second. 
{\bf Ans.} 27/14 ft. per sec.  Since $s(1) = 27/14$ and $s(0) = 0$, the distance travelled is
27/14 ft.  The time elaped is 1 sec. \  (b)  What is the acceleration of the car after $4$
seconds? {\bf Ans.} 20 ft. per ${\rm sec}^2$.  Compute $s''$ and plug in $t=4$.
  
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