Mathematics 265 Introduction to Calculus I

Solutions to Sample Midterm Exam

Sample Midterm Examination 2

Time: 3 hours
Passing grade: 55%
Total points: 64

Solutions and Marking Scheme

These are our solutions. Your solutions may be different but equally correct. If you have any doubts, consult your tutor. Full credit is given if

• the question is answered: you find what we are asking for, you describe what we are asking for, and you gave what we are asking for.
• the answer is correct.
• you explain and/or justify each of your steps.

1. (4 points)

1. Determine which of the following equations are well defined functions with an independent variable ${\displaystyle x}$. Explain.

   1. ${\displaystyle {\displaystyle \frac{2y}{x^2-3\left|x\right|}}=1}$

   2. ${\displaystyle y^2+x^2=10}$

   Solution

   1. If we solve for ${\displaystyle y}$, we get ${\displaystyle y={\displaystyle \frac{x^2-\left|x\right|}{2}}}$. (1 pt)

      Hence, this is an algebraic function and ${\displaystyle \left|x\right|}$ is well defined for any ${\displaystyle x}$, therefore for every ${\displaystyle x}$ there is one and only one value of ${\displaystyle y}$ and the function is well defined. (1 pt)

   2. ${\displaystyle y^2+x^2=10}$ it is not, (1 pt)

      because ${\displaystyle \left(-\sqrt{10},0\right)}$ and ${\displaystyle \left(\sqrt{10},0\right)}$ satisfy the equation. (1 pt)

2. (4 points)
2. Consider a cone whose height is double the size of its radius.
   1. Express the volume ${\displaystyle V}$ of the cone as a function of radius ${\displaystyle r}$.
   2. Find the domain of the function in a.
   3. Use the function in a. to compute the volume of the cone for ${\displaystyle r=3}$.

   Hint: The volume of a cone is ${\displaystyle V={\displaystyle \frac{\pi r^{2}h}{3}}}$.

   Solution

   1. Since ${\displaystyle h=2r}$, (1 pt)

      we have ${\displaystyle V={\displaystyle \frac{\pi r^2\left(2r\right)}{3}}={\displaystyle \frac{2\pi r^3}{3}}}$. (1 pt)

   2. Since ${\displaystyle r}$ is the radius, the domain is ${\displaystyle \left[0,\infty \right)}$. (1 pt)

   3. ${\displaystyle V\left(3\right)={\displaystyle \frac{2\pi 3^3}{3}}={\displaystyle \frac{54\pi }{3}}}$. (1 pt)

3. (4 points)
3. The population ${\displaystyle P}$ of a city in the year ${\displaystyle y}$ is ${\displaystyle P\left(y\right)}$. Write a sentence (in layman’s terms) that explains the meaning of the expressions
   1. ${\displaystyle {\displaystyle \frac{p\left(15\right)-P\left(2\right)}{13}}=1\text{,}431}$
   2. ${\displaystyle P^{\prime }\left(10\right)=14\text{,}000}$.

   Solution

   1. ${\displaystyle {\displaystyle \frac{p\left(15\right)-P\left(2\right)}{13}}={\displaystyle \frac{p\left(15\right)-P\left(2\right)}{15-2}}=1431}$ this is the average on the interval ${\displaystyle [2,15]}$, (1 pt)

      hence from the years 2 to 15, on average the population is increasing at a rate of 1,431 people per year. (1 pt)

   2. ${\displaystyle P^{\prime }\left(10\right)=14\text{,}000}$ is an instant rate of change. (1 pt)

      Hence, in the 10th year, the population is increasing at a rate of 14,000 people per year. (1 pt)

4. (6 points)

4. Indicate the transformations in the order they are applied to the basic graph of ${\displaystyle \left|x\right|}$ in the interval in order to obtain the graph of the function ${\displaystyle f\left(x\right)=-3|x-2|}$. Apply the transformations to sketch the graph of ${\displaystyle f}$, no credit will be given if other method is used.

   Solution

   1. Shift to the right by 2 units gives ${\displaystyle |x-2|}$. (1 pt)

   2. Reflection with respect to the ${\displaystyle x}$-axis gives ${\displaystyle -|x-2|}$. (1 pt)

   3. Vertical stretch by 3 units gives ${\displaystyle f\left(x\right)=-3|x-2|}$. (1 pt)

   The graph is

   (3 pts)

5. (10 points)
5. Interpret in terms of the graph of the function ${\displaystyle f}$ the conditions listed below. Sketch the graph of a single function ${\displaystyle f\left(x\right)}$ which satisfies all of the conditions listed below.
   1. ${\displaystyle \underset{x\to \infty }{\lim }f(x)=3}$
   2. ${\displaystyle $ f(0)=0$}$
   3. ${\displaystyle \underset{x\to -1}{\lim }f(x)=2}$
   4. ${\displaystyle \underset{x\to 1}{\lim }f\left(x\right)=\infty }$
   5. the function is continuous but not differentiable at ${\displaystyle x=0}$

   Solution

   1. ${\displaystyle \underset{x\to \infty }{\lim }f(x)=3y=3}$ is a horizontal asymptote in the positive direction. (1 pt)

   2. ${\displaystyle f\left(0\right)=0}$: the graph passes through the point ${\displaystyle (0,0)}$. (1 pt)

   3. ${\displaystyle \underset{x\to -1}{\lim }f(x)=2}$: the values of the function from the right and left at ${\displaystyle -1}$ are close to ${\displaystyle 2}$. (2 pts)

   4. ${\displaystyle \underset{x\to 1}{\lim }f(x)=\infty }$ is a vertical asymptote in the positive direction from the left and right of $1$. (2 pts)

   5. the function is continuous but not differentiable at ${\displaystyle x=0}$: the graph has a kink or vertical tangent line at ${\displaystyle (0,0)}$. (1 pt)

   (3 pts)

6. (10 points)

6. Find the vertical and horizontal asymptotes of the function

   ${\displaystyle f\left(x\right)={\displaystyle \frac{x^2-3x-4}{x^2-16}}.}$

   Solution

   1. Vertical Asymptotes.

      ${\displaystyle x^2-16=\left(x-4\right)\left(x+4\right)}$ possible asymptotes ${\displaystyle x=4,x=-4}$. (1 pt)

      ${\displaystyle {\displaystyle \frac{x^2-3x-4}{x^2-16}}={\displaystyle \frac{\left(x+1\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}}={\displaystyle \frac{x+1}{x+4}}}$ (1 pt)

      ${\displaystyle \underset{x\to -4^{+}}{\lim }{\displaystyle \frac{{\left(x^2-3x-4\right)}^2}{x^2-16}}=\underset{x\to -4^{+}}{\lim }{\displaystyle \frac{x+1}{x+4}}=-\infty }$ (1 pt)

      because ${\displaystyle \underset{x\to -4}{\lim }x+1=-3<0}$, ${\displaystyle x+4>0}$ for ${\displaystyle x\to -4^{+}}$ and ${\displaystyle \underset{x\to -4^{+}}{\lim }x+4=0}$. (3 pts)

      The only vertical asymptote is ${\displaystyle x=-4}$. (1 pt)

   2. Horizontal Asymptotes.

      ${\displaystyle \underset{x\to \infty }{\lim }{\displaystyle \frac{{\left(x^2-3x-4\right)}^2}{x^2-16}}=1}$

      ${\displaystyle \underset{x\to -\infty }{\lim }{\displaystyle \frac{{\left(x^2-3x-4\right)}^2}{x^2-16}}=1}$ (1 pt)

      because the degrees of the numerator an denominator are equal. (1 pt)

      So ${\displaystyle y=1}$ is the only horizontal asymptote. (1 pt)

7. (24 points)

7. Evaluate each of the limits given below. If a limit does not exist, explain why.

   Note: No credit will be given for unjustified answers.

   1. ${\displaystyle \underset{x\to 1}{lim}{\displaystyle \frac{x^2-1}{x^3-1}}}$
   2. ${\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{x^2}{1-\cos \left(2x\right)}}}$
   3. ${\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{6x-9}{x^3-12x+3}}}$
   4. ${\displaystyle \underset{x\to \infty }{lim}{\displaystyle \frac{5-2x^3}{x^2+2}}}$
   5. ${\displaystyle \underset{x\to -\pi /3}{lim}{\displaystyle \frac{\tan \left(2x\right)}{3x+\pi }}}$
   6. ${\displaystyle \underset{x\to 2}{lim}{\displaystyle \frac{\cos \left(\pi x\right)}{{\left(x-2\right)}^2}}}$

   Solution

   1. ${\displaystyle {\displaystyle \frac{x^2-1}{x^3-1}}={\displaystyle \frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}}={\displaystyle \frac{x+1}{x^2+x+1}}}$ (2 pts)

      ${\displaystyle \underset{x\to 1}{\lim }{\displaystyle \frac{x^2-1}{x^3-1}}=\underset{x\to 1}{\lim }{\displaystyle \frac{x+1}{x^2+x+1}}={\displaystyle \frac{2}{3}}}$ (2 pts)

   2. We use the identity ${\displaystyle 1-{\cos }^{2}a={\sin }^{2}a}$

      ${\displaystyle {\displaystyle \frac{x^2}{1-\cos (2x)}}={\displaystyle \frac{x^2}{1-\cos (2x)}}\left({\displaystyle \frac{1+\cos (2x)}{1+\cos (2x)}}\right)}$

      ${\displaystyle \phantom{{\displaystyle \frac{x^2}{1-\cos (2x)}}}={\displaystyle \frac{x^2\left(1+\cos (2x)\right)}{1-{\cos }^2(2x)}}}$

      ${\displaystyle \phantom{{\displaystyle \frac{x^2}{1-\cos (2x)}}}={\displaystyle \frac{x^2\left(1+\cos (2x)\right)}{{\sin }^2(2x)}}}$

      ${\displaystyle \phantom{{\displaystyle \frac{x^2}{1-\cos (2x)}}}={\left({\displaystyle \frac{x}{\sin (2x)}}\right)}^2\left(1+\cos (2x)\right)}$ (2 pts)

      By the law of limits and trigonometric limit ${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{x}{\sin (2x)}}={\displaystyle \frac{1}{2}}}$

      ${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{x^2}{1-\cos \left(2x\right)}}=\underset{x\to 0}{\lim }{\left({\displaystyle \frac{x}{\sin \left(2x\right)}}\right)}^2\left(1+\cos \left(2x\right)\right)={\displaystyle \frac{2}{4}}={\displaystyle \frac{1}{2}}}$ (2 pts)

   3. The function ${\displaystyle {\displaystyle \frac{6x-9}{x^3-12x+3}}}$ is continuous at zero because it is rational and the denominator is nonzero at ${\displaystyle 0}$. (1 pt)

      Hence,

      ${\displaystyle \underset{x\to 0}{lim}{\displaystyle \frac{6x-9}{x^3-12x+3}}={\displaystyle \frac{-9}{3}}=-3}$ (1 pt)

   4. The function ${\displaystyle {\displaystyle \frac{5-2x^3}{x^2+2}}}$ is rational and the degree of the numerator (3) is greater than the degree of the numerator (2). Hence the limit at infinity may be infinity or negative infinity. (2 pts)

      ${\displaystyle \underset{x\to \infty }{\lim }{\displaystyle \frac{5-2x^3}{x^2+2}}=\underset{x\to \infty }{\lim }{\displaystyle \frac{-2x^3}{x^2}}=\underset{x\to \infty }{\lim }-2x=-\infty }$ (2 pts)

   5. The function is not continuous at ${\displaystyle -\pi /3}$, and ${\displaystyle \tan \left(-{\displaystyle \frac{2\pi }{3}}\right)=-\sqrt{3}<0}$. (1 pt)

      ${\displaystyle \underset{x\to -\pi /3^{-}}{\lim }{\displaystyle \frac{1}{3x+\pi }}=-\infty }$   and   ${\displaystyle \underset{x\to -\pi /3^{+}}{\lim }{\displaystyle \frac{1}{3x+\pi }}=\infty }$

      because ${\displaystyle 3x+\pi <0}$ for ${\displaystyle x\to -\pi /3^{-}}$ and ${\displaystyle 3x+\pi >0}$ for ${\displaystyle x\to -\pi /3^{+}}$ (3 pts)

      Hence

      ${\displaystyle \underset{x\to -\pi /3^{-}}{\lim }{\displaystyle \frac{\tan (2x)}{3x+\pi }}=\infty }$   and   ${\displaystyle \underset{x\to -\pi /3^{+}}{\lim }{\displaystyle \frac{\tan (2x)}{3x+\pi }}=-\infty }$

      so the limit does not exist. (2 pts)

   6. The function is not continuous at 2.

      ${\displaystyle \underset{x\to 2}{\lim }{\displaystyle \frac{1}{{\left(x-2\right)}^2}}=\infty }$

      because ${\displaystyle {\left(x-2\right)}^2>0}$ for any ${\displaystyle x\ne 2}$, and ${\displaystyle \underset{x\to 2}{\lim }{\left(x-2\right)}^2=0}$ (2 pts)

      Since

      ${\displaystyle \underset{x\to 2}{\lim }\cos \left(\pi x\right)=\cos \left(2\pi \right)=1>0}$

      we conclude that

      ${\displaystyle \underset{x\to 2}{\lim }{\displaystyle \frac{\cos \left(\pi x\right)}{{\left(x-2\right)}^2}}=\infty }$ (2 pts)

8. (6 points)

8. Use linearization to estimate the value of ${\displaystyle \sin (62°)}$.

   Solution

   We set ${\displaystyle f(x)=\sin x}$, ${\displaystyle a=60°={\displaystyle \frac{\pi }{3}}}$, ${\displaystyle \Delta x=2°={\displaystyle \frac{\pi }{90}}}$. (2 pts)

   ${\displaystyle f(a+\Delta x)\approx f^{\prime }(a)\Delta x+f(a)=\cos \left({\displaystyle \frac{\pi }{3}}\right)\left({\displaystyle \frac{\pi }{90}}\right)+\sin \left({\displaystyle \frac{\pi }{3}}\right)}$ (2 pts)

   We have

   ${\displaystyle \sin (62°)\approx {\displaystyle \frac{\pi }{180}}+{\displaystyle \frac{\sqrt{3}}{2}}}$ (2 pts)

9. (16 points)

9. Find each of the derivatives listed below. Show your work.

   Note: You may not need to simplify your answer.

   1. ${\displaystyle {\displaystyle \frac{d^2}{d{x}^2}}\cot \left(2x\right)}$

   2. ${\displaystyle {\displaystyle \frac{d}{dx}}\sec \left(x^2-3x\right)}$

   3. ${\displaystyle {\displaystyle \frac{d}{dx}}{\displaystyle \frac{x^2-\cos \left(3x\right)}{x\sin \left(2x\right)}}}$

   4. ${\displaystyle {\displaystyle \frac{d}{dx}}{{\displaystyle \frac{f(x)}{g(x)}}|}_{x=1}}$ where ${\displaystyle f\left(1\right)=3}$, ${\displaystyle g\left(1\right)=6}$, ${\displaystyle f^{\prime }\left(x\right)=\sqrt{x}}$ and ${\displaystyle g^{\prime }\left(x\right)=3\sin \left(\pi x/3\right)}$.

   Solution

   1. ${\displaystyle {\displaystyle \frac{d}{dx}}\cot \left(2x\right)=-2{\csc }^2\left(2x\right)}$ (1 pt)

      ${\displaystyle {\displaystyle \frac{d^2}{d{x}^2}}\cot \left(2x\right)={\displaystyle \frac{d}{dx}}-2{\csc }^2\left(2x\right)}$

      ${\displaystyle \phantom{{\displaystyle \frac{d^2}{d{x}^2}}\cot \left(2x\right)}=-2\left(-2\csc \left(2x\right)\cot \left(2x\right)\csc \left(2x\right)\left(2\right)\right)}$

      ${\displaystyle \phantom{{\displaystyle \frac{d^2}{d{x}^2}}\cot \left(2x\right)}=8{\csc }^2\left(2x\right)\cot \left(2x\right)}$ (2 pts)

   2. ${\displaystyle {\displaystyle \frac{d}{dx}}\sec \left(x^2-3x\right)=\sec \left(x^2-3x\right)\tan \left(x^2-3x\right)\left(2x-3\right)}$ (2 pts)

   3. ${\displaystyle f(x)=x^2-\cos (3x)}$ and ${\displaystyle g(x)=x\sin (2x)}$.

      Thus, ${\displaystyle f^{\prime }(x)=2x+3\sin (3x)}$ and ${\displaystyle g^{\prime }(x)=\sin \left(2x\right)+2x\cos \left(2x\right)}$. (2 pts)

      ${\displaystyle {\displaystyle \frac{d}{dx}}{\displaystyle \frac{x^2-\cos \left(3x\right)}{x\sin \left(2x\right)}}}$

      $$\begin{gathered} {\displaystyle ={\displaystyle \frac{x\sin \left(2x\right)\left[2x+3\sin \left(3x\right)\right]- \\ \left(x^2-\cos \left(3x\right)\right)\left[\sin \left(2x\right)+2x\cos \left(2x\right)\right]}{x^2{\sin }^2\left(2x\right)}}} \end{gathered}$$ (2 pts)

   4. By the quotient rule

      ${\displaystyle {\displaystyle \frac{d}{dx}}{{\displaystyle \frac{f(x)}{g(x)}}|}_{x=1}={{\displaystyle \frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{g{(x)}^2}}|}_{x=1}}$

      ${\displaystyle \phantom{{\displaystyle \frac{d}{dx}}{{\displaystyle \frac{f(x)}{g(x)}}|}_{x=1}}={{\displaystyle \frac{f^{\prime }(1)g(1)-f(1)g^{\prime }(1)}{g{(1)}^2}}|}_{x=1}}$ (2 pts)

      ${\displaystyle f^{\prime }(1)=\sqrt{1}=1}$ and ${\displaystyle g^{\prime }(1)=3\sin (\pi /3)={\displaystyle \frac{3\sqrt{3}}{2}}}$. (1 pt)

      Thus,

      ${\displaystyle \frac{d}{dx}}{\displaystyle \frac{f(x)}{g(x)}}{|}_{x=1}={\displaystyle \frac{6-{\displaystyle \frac{9\sqrt{3}}{2}}}{36}}={\displaystyle \frac{2-3\sqrt{3}}{12}}$ (2 pts)

10. (4 points)

10. Find the equation of the tangent line to the curve ${\displaystyle y^3+y{x}^2+x^2=3y^2}$ at the point ${\displaystyle (1,1)}$.

    Solution

    By implicit differentiation

    ${\displaystyle 3y^2{y}^{\prime }+x^2{y}^{\prime }+2xy+2x=6y{y}^{\prime }}$   for ${\displaystyle x=1}$ and ${\displaystyle y=1}$   ${\displaystyle 4y^{\prime }+4=6y^{\prime }}$ (2 pts)

    Solving for ${\displaystyle y^{\prime }=2}$. The equation of the line is ${\displaystyle y-1=2(x-1)}$ or ${\displaystyle y=2x-1}$. (2 pts)

11. (6 points)

11. A rocket, rising vertically, is tracked by a radar station that is on the ground 5 mi from the launch pad. How fast is the rocket rising when it is 4 mi high and its distance from the radar station is increasing at a rate of 2000 mi/h?

    Solution

    (2 pts)

    According to the diagram, we want to know ${\displaystyle {\displaystyle \frac{dx}{dt}}}$ when ${\displaystyle x=4}$ and ${\displaystyle {\displaystyle \frac{dz}{dt}}=2000}$. (1 pt)

    By Pythagoras theorem, ${\displaystyle z^2=25+x^2}$ and ${\displaystyle 2z{\displaystyle \frac{dz}{dt}}=2x{\displaystyle \frac{dx}{dt}}}$ , so for ${\displaystyle x=4}$, we have ${\displaystyle z=\sqrt{41}}$; hence, (2 pts)

    ${\displaystyle 2\sqrt{41}\left(2000\right)=2(4){\displaystyle \frac{dx}{dt}}}$   ${\displaystyle {\displaystyle \frac{dx}{dt}}=500\sqrt{41}\text{\hspace{0.17em} mi/hr}}$ (1 pt)