Fall 21 Final Solution

Problems

  1. Determine the following limits. If you apply L'Hopital's rule, indicate where you have applied it and why you can apply it. If your final answer is "does not exist," \(\displaystyle \infty\), or \(\displaystyle -\infty\), briefly explain your answer. (You will not receive full credit for a "does not exist" answer if the answer is \(\displaystyle \infty\) or \(\displaystyle -\infty\).)

    1. \(\displaystyle \lim_{x \to 5} (4 + 7x - x^2)\)
    2. \(\displaystyle \lim_{x \to 0} \frac{e^{3x} - 1}{7x}\)
    3. \(\displaystyle \lim_{x \to \infty} \frac{7x^5 - 3x^2 + 1}{2x^5 + x^2 - 6}\)

    Solution:

    1. This is a polynomial, and polynomials are continuous everywhere, so we may substitute directly: \[ 4+7(5)-5^2=4+35-25=14. \] Therefore \[ \boxed{14}. \]
    2. As \(x\to0\), we use the standard limit \[ \frac{e^{3x}-1}{3x}\to1. \] Rewrite the given expression as \[ \frac{e^{3x}-1}{7x}=\frac{3}{7}\cdot\frac{e^{3x}-1}{3x}. \] Taking the limit gives \[ \boxed{\frac37}. \]
    3. For a rational function with the same highest degree in numerator and denominator, the limit at infinity is the ratio of the leading coefficients. Here the leading terms are \(7x^5\) and \(2x^5\), so \[ \lim_{x\to\infty}\frac{7x^5-3x^2+1}{2x^5+x^2-6}=\frac{7}{2}. \] Thus \[ \boxed{\frac72}. \]

  2. Consider the graph of \(\displaystyle y = f(x)\) given below. The grid lines are one unit apart. Based on the graph answer the following questions.

    y

    x

    1. Determine \(\displaystyle \lim_{x \to 2} f(x)\).
    2. Determine \(\displaystyle f(0)\).
    3. Determine all values of \(\displaystyle x\) in the interval \(\displaystyle (-5, 4)\) at which \(\displaystyle f(x)\) is NOT continuous. Briefly explain your thinking.
    4. Determine all values of \(\displaystyle x\) in the interval \(\displaystyle (-5, 4)\) at which \(\displaystyle f'(x)\) is undefined. Briefly explain your thinking.

    IMAGE

    Graph of the derivative of a function

    Solution:

    1. To find \(\lim_{x\to2}f(x)\), we look at the y-value the graph approaches from both sides of \(x=2\). From the picture, both sides approach the same height, namely \(2\). Therefore \[ \boxed{\lim_{x\to2}f(x)=2}. \]
    2. The value \(f(0)\) is read from the filled point on the graph at \(x=0\). That filled point has height \(2\), so \[ \boxed{f(0)=2}. \]
    3. The function is not continuous at points where the graph has a hole, jump, or mismatch between the limiting value and the actual function value. From the graph, these issues occur at \(x=-4\) and \(x=0\). Therefore the points of discontinuity in \((-5,4)\) are \[ \boxed{x=-4 \text{ and } x=0}. \]
    4. The derivative is undefined at any point where the function is discontinuous, and also at sharp corners or cusps. From the graph, this happens at the discontinuities \(x=-4\) and \(x=0\), and at the sharp corner at \(x=2\). Hence \[ \boxed{x=-4,\ 0,\ 2}. \]

  3. (a) State the limit definition of the derivative of \(\displaystyle f(x)\).

    1. Use the \textbf{limit definition of the derivative} to determine the derivative of \(\displaystyle f(x) = \sqrt{8 - 3x}\). No points will be awarded for the application of differentiation rules (and L'Hopital’s rule is not allowed).

    Solution: The limit definition is \[ f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}. \] For \[ f(x)=\sqrt{8-3x}, \] the difference quotient is \[ \frac{\sqrt{8-3x-3h}-\sqrt{8-3x}}{h}. \] To simplify it, rationalize the numerator by multiplying by the conjugate: \[ \frac{\sqrt{8-3x-3h}-\sqrt{8-3x}}{h}\cdot \frac{\sqrt{8-3x-3h}+\sqrt{8-3x}}{\sqrt{8-3x-3h}+\sqrt{8-3x}}. \] The numerator becomes \[ (8-3x-3h)-(8-3x)=-3h, \] so the quotient simplifies to \[ \frac{-3}{\sqrt{8-3x-3h}+\sqrt{8-3x}}. \] Now let \(h\to0\): \[ f'(x)=\lim_{h\to0}\frac{-3}{\sqrt{8-3x-3h}+\sqrt{8-3x}} =-\frac{3}{2\sqrt{8-3x}}. \] Therefore \[ \boxed{f'(x)=-\frac{3}{2\sqrt{8-3x}}}. \]

  4. Determine the first derivative of each of the following functions. Remember to use correct notation to write your final answer.

    1. \(\displaystyle f(x) = e^{x^2} + \frac{5}{x^2} + \frac{x}{x+1}\)
    2. \(\displaystyle f(x) = (2x^5 + 3) \tan(x)\)
    3. \(\displaystyle f(x) = (\arcsin(3x))^2\)
    4. \(\displaystyle y = x^{3x^2-x}\)

    Solution:

    1. \(f'(x)=\boxed{2xe^{x^2}-\frac{10}{x^3}+\frac{1}{(x+1)^2}}\).
    2. \(f'(x)=\boxed{10x^4\tan x+(2x^5+3)\sec^2 x}\).
    3. \(f'(x)=\boxed{\frac{6\arcsin(3x)}{\sqrt{1-9x^2}}}\).
    4. For \(y=x^{3x^2-x}\), \(\ln y=(3x^2-x)\ln x\), so \(\boxed{y'=x^{3x^2-x}\left((6x-1)\ln x+3x-1\right)}\).

  5. Determine the following indefinite integrals.

    1. \(\displaystyle \int \left( \frac{6x^4 + 3x^3 - 8}{x} \right) dx\)
    2. \(\displaystyle \int \left( \cos(2x) - \frac{1}{1+x^2} \right) dx\)
    3. \(\displaystyle \int \frac{1}{\sqrt{x}(5+2\sqrt{x})^3} dx\)

    Solution:

    1. \(\displaystyle \int \frac{6x^4+3x^3-8}{x}dx=\int(6x^3+3x^2-8/x)dx=\boxed{\frac32x^4+x^3-8\ln|x|+C}\).
    2. \(\displaystyle \int\left(\cos(2x)-\frac{1}{1+x^2}\right)dx=\boxed{\frac12\sin(2x)-\arctan x+C}\).
    3. Let \(u=\sqrt{x}\), so \(x=u^2\), \(dx=2u\,du\). Then the integral becomes \(2\int(5+2u)^{-3}du\), giving \(\boxed{-\frac{1}{2(5+2\sqrt{x})^2}+C}\).

  6. Evaluate the following definite integrals.

    1. \(\displaystyle \int_{0}^{1} (x^3 + e^{-3x}) dx\)
    2. \(\displaystyle \int_{0}^{4} \sqrt{4y + 9} dy\)
    3. \(\displaystyle \int_{0}^{\pi/2} \frac{\sin x}{1 + \cos x} dx\)

    Solution:

    1. Evaluate term by term: \[ \int_0^1(x^3+e^{-3x})\,dx=\int_0^1x^3\,dx+\int_0^1e^{-3x}\,dx. \] An antiderivative is \[ \frac{x^4}{4}-\frac13e^{-3x}. \] So \[ \left[\frac{x^4}{4}-\frac13e^{-3x}\right]_0^1 =\left(\frac14-\frac13e^{-3}\right)-\left(0-\frac13\right) =\frac{7-e^{-3}}{12}. \] Therefore \[ \boxed{\frac{7-e^{-3}}{12}}. \]
    2. Use the substitution \(u=4y+9\), so \(du=4\,dy\) and \(dy=\dfrac14du\). When \(y=0\), \(u=9\); when \(y=4\), \(u=25\). Then \[ \int_0^4\sqrt{4y+9}\,dy =\frac14\int_9^{25}u^{1/2}\,du =\frac14\cdot\frac{2}{3}u^{3/2}\Big|_9^{25}. \] Thus \[ \frac16\left(25^{3/2}-9^{3/2}\right)=\frac16(125-27)=\frac{98}{6}=\frac{49}{3}. \] So the value is \[ \boxed{\frac{49}{3}}. \]
    3. Use the substitution \[ u=1+\cos x,\qquad du=-\sin x\,dx. \] When \(x=0\), \(u=2\); when \(x=\pi/2\), \(u=1\). Therefore \[ \int_0^{\pi/2}\frac{\sin x}{1+\cos x}\,dx =-\int_2^1\frac{1}{u}\,du =\int_1^2\frac{1}{u}\,du =\ln 2. \] Hence \[ \boxed{\ln 2}. \]

  7. Water is added to an empty rain barrel at a rate of \(\displaystyle 30 - 2t\) gallons per hour, starting at time \(\displaystyle t = 0\), until the tank is completely full. If the rain barrel holds 225 gallons, how long will it take to completely fill the tank?

    Hint: What does \(\displaystyle 30 - 2t\) gallons per hour represent?

    Solution: The expression \(30-2t\) gallons per hour is the rate at which water is entering the barrel. Therefore, the total amount of water added from time \(0\) to time \(T\) is \[ \int_0^T(30-2t)\,dt. \] Compute the integral: \[ \int_0^T(30-2t)\,dt=\left[30t-t^2\right]_0^T=30T-T^2. \] The barrel is full when this total added amount equals \(225\) gallons, so we solve \[ 30T-T^2=225. \] Rearrange: \[ T^2-30T+225=0. \] Factor: \[ (T-15)^2=0. \] Thus \[ \boxed{T=15}. \] So it takes \(\boxed{15\text{ hours}}\) to fill the barrel completely.

  8. The surface area of a cube of ice is decreasing at a rate of \(\displaystyle 10 \ \mathrm{cm}^2/\mathrm{s}\). At what rate is the volume of the cube changing when the surface area is \(\displaystyle 24 \ \mathrm{cm}^2\)?

    Solution: Let \(s\) be the side length of the cube. Then \[ A=6s^2,\qquad V=s^3. \] We are given \[ \frac{dA}{dt}=-10. \] At the instant of interest, the surface area is \(24\), so \[ 6s^2=24 \quad\Longrightarrow\quad s^2=4 \quad\Longrightarrow\quad s=2 \] since side length is positive.

    Differentiate the surface-area formula: \[ \frac{dA}{dt}=12s\frac{ds}{dt}. \] Substitute \(s=2\) and \(\dfrac{dA}{dt}=-10\): \[ -10=12(2)\frac{ds}{dt}=24\frac{ds}{dt}. \] So \[ \frac{ds}{dt}=-\frac{5}{12}. \] Now differentiate the volume formula: \[ \frac{dV}{dt}=3s^2\frac{ds}{dt}. \] Substitute \(s=2\) and \(\dfrac{ds}{dt}=-\dfrac{5}{12}\): \[ \frac{dV}{dt}=3(4)\left(-\frac{5}{12}\right)=-5. \] Therefore \[ \boxed{\frac{dV}{dt}=-5\text{ cm}^3/\text{s}}. \]

  9. A rectangle is to be inscribed in the ellipse \(\displaystyle \frac{x^2}{4} + y^2 = 1\). (See diagram below.)

    (Diagram of an ellipse with axes labeled \(\displaystyle x\) and \(\displaystyle y\), and a rectangle inscribed. The top-right corner is marked as \(\displaystyle (x, y)\).)

    1. Let \(\displaystyle x\) represent the \(\displaystyle x\)-coordinate of the top-right corner of the rectangle. Determine a formula \(\displaystyle A(x)\) for the area of the rectangle as a function of \(\displaystyle x\) alone. Your final answer should include the variable \(\displaystyle x\) and can not include any other variables.
    2. Suppose you want to determine the dimensions of the rectangle that will result in the maximum possible area. Determine an appropriate domain for \(\displaystyle A(x)\). Briefly explain your answer. (Note: We will not actually maximize \(\displaystyle A(x)\) in this problem.)

    IMAGE

    Diagram of an ellipse with an inscribed rectangle

    Solution:

    1. In the ellipse \(x^2/4+y^2=1\), the top-right corner has coordinates \((x,y)\), so \(y=\sqrt{1-x^2/4}\). The rectangle has width \(2x\) and height \(2y\), hence area \(A(x)=4xy=\boxed{4x\sqrt{1-x^2/4}}\).
    2. A reasonable domain is \(0\le x\le 2\), since \(x\) is the positive x-coordinate of the top-right corner and must remain inside the ellipse.

  10. Suppose you want to find the dimensions (in centimeters) of the open-top cylinder with volume \(\displaystyle V = 16\pi\) cubic centimeters that has the minimum possible surface area.

    The surface area of an open-top cylinder with radius \(\displaystyle r\) centimeters and height \(\displaystyle h\) centimeters is \(\displaystyle S = 2\pi rh + \pi r^2\) square centimeters. Since the volume is \(\displaystyle 16\pi\) cubic centimeters, we know that \(\displaystyle 16\pi = \pi r^2 h\).

    1. Determine the surface area \(\displaystyle S\) as a function of \(\displaystyle r\). (The variable \(\displaystyle h\) should not be used in your answer.)
    2. Determine an appropriate domain for the surface area function \(\displaystyle S\) from part (a), and explain briefly.
    3. Use calculus techniques to determine the value of \(\displaystyle r\) which results in the minimum possible surface area. Write a sentence, using appropriate units, to summarize your answer.

    Solution:

    1. From \(16\pi=\pi r^2h\), we get \(h=16/r^2\). Hence \(S(r)=2\pi r\cdot 16/r^2+\pi r^2=\boxed{\frac{32\pi}{r}+\pi r^2}\).
    2. The domain is \(r>0\), because a radius must be positive.
    3. Differentiate: \(S'(r)=-32\pi/r^2+2\pi r\). Setting this to zero gives \(2\pi r^3=32\pi\), so \(r^3=16\) and \(r=\boxed{\sqrt[3]{16}\text{ cm}}\). Then \(h=16/r^2=\boxed{\sqrt[3]{16}\text{ cm}}\) as well. Since \(S''(r)=64\pi/r^3+2\pi>0\), this is the minimum surface area.