Math Insight

Approximating a nonlinear function by a linear function

Math 201, Spring 2016
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Due date: Oct. 9, 2015, 11:59 p.m.
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Total points: 3
  1. Let $g(x) = x^3$. Find the average rate of change in $g(x)$ between:

    1. $x=3$ and $x=4$.

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    2. $x=3$ and $x=3.1$.
    3. $x=3$ and $x=3.01$.
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    4. $x=2$ and $x=3$.
    5. $x=2.9$ and $x=3$.
    6. $x=2.99$ and $x=3$.
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    7. $x=a$ and $x=b$
    8. $x=a$ and $x=a+h$

  2. Feedback from applet
    number of secants:
    number of tangents:
    secants:
    tangents:
    Let $f(x) = - x^{2} + 10 x$, which is graphed to the right.
    1. Sketch a secant line from $x=2$ to $x=4$. What is its slope?
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    2. Sketch a tangent line at $x=2$. Estimate its slope by calculating the slope of secant lines from $x=2$ to $x=2+h$ for the following values of $h$:
      • $h=0.1$:
      • $h= 0.01$:
      • $h= 0.001$:
      • $h=-0.1$:
      • $h= -0.01$:
      • $h= -0.001$:
      (Round above answers to nearest 1000th or less.)

      As $h$ approach zero, it appears the slope approaches what value?

      Based on this observation, estimate the the slope of the tangent line at $x=2$:
      $f'(2) = $
      .

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    3. Sketch a tangent line at $x=4$. Estimate its slope by calculating the slope of secant lines from $x=4$ to $x=4+h$ for the following values of $h$:
      • $h=0.1$:
      • $h= 0.01$:
      • $h= 0.001$:
      • $h=-0.1$:
      • $h= -0.01$:
      • $h= -0.001$:
      (Round above answers to nearest 1000th or less.)

      The slope of the tangent line appears to be $\displaystyle\diff{f}{x}\bigg|_{x=4} =$
      .

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  3. A squirrel is running along a branch avoiding an oncoming owl. Her position $x$ along the branch after $t$ seconds is $g(t)=4 t^{5} - 52 t^{4} + 236 t^{3} - 424 t^{2} + 240 t$, where $g(t)$ is given in cm, relative to her position at time $t=0$ seconds. The graph of $x=g(t)$ is shown below.
    Feedback from applet
    number of secants:
    number of tangents:
    secants:
    tangents:
    1. After five seconds, her position along the branch is $g(5)=100$ cm. What is the average rate of change of her position (i.e., her average velocity) during the five seconds from $t=0$ to $t=5$?
      cm/second.

      Sketch a secant line on the above graph that shows her position as a function of time if she had moved steadily at that velocity during those five seconds. What is the equation of that line?
      $x=$

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    2. Her position at $t=1$ seconds is $g(1)=4$ cm; at $t=2$ seconds, it is $g(2)=-32$ cm. What is her average velocity during the period between $1 \le t \le 2$?
      cm/second.

      Sketch a secant line on the above graph, which would show her position if she had moved steadily at that velocity starting at $t=1$. What is the equation of that line?
      $x=$

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    3. Approximate the instantaneous rate of her position (her velocity) at $t=3$ seconds. To do so, calculate her average rate of change from $t=3$ to $t=3+\Delta t$ for $\Delta t=0.01$ and $0.0001$.

      Estimated velocity from $\Delta t=0.01$:
      cm/second.
      Estimated velocity from $\Delta t=0.001$:
      cm/second.
      (Round the above answers to nearest 100th of a cm/second or less.)

      As $\Delta t$ decreases, it appears the velocity is approaching what integer value?
      cm/second

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    4. Assuming she continued moving steadily at this velocity starting at $t=3$ seconds, write down an equation for her position as a function of time.
      $x=$

      The graph of this equation will be a tangent line to the graph at $t=3$ seconds. Sketch the tangent line on the above graph.

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  4. The slope of a graph at a point is the same thing as the slope of the tangent line at that point. Estimate the slope of the function $f(x)=x^{2}$ at $x=1$ by calculating the slope of the secant line between $x=1$ and $x=1+\Delta x$ for the following values of $\Delta x$:
    • $\Delta x = 0.1$:
    • $\Delta x = 0.01$:
    • $\Delta x = 0.001$:
    • $\Delta x = -0.1$:
    • $\Delta x = -0.01$:
    • $\Delta x = -0.001$:
    (Round answer to nearest 1000th or less.)

    Since for both positive $\Delta x$ and negative $\Delta x$, the slope of the secant is approaching the same value as $\Delta x$ approaches zero, the slope of the function at $x=1$ must be
    .

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Math 201, Spring 2016