Approximating a nonlinear function by a linear function

Math 1241, Fall 2020
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Due date: Oct. 9, 2020, 11:59 p.m.
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Total points: 3

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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  Hint
  The average rate of change is the slope of the secant line, i.e., the slope of the line from the point $(3,3^3)$ to the point $(4,4^3)$.
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  Approximating a nonlinear function by a linear function
  
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2. $x=3$ and $x=3.1$.
3. $x=3$ and $x=3.01$.
  
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  Hint
  You can round your answer to the nearest 0.01.
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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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  Hint
  You can round your answer to the nearest 0.01.
  If all went well, your answer should be close to your answer for part (c).
  
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7. $x=a$ and $x=b$
8. $x=a$ and $x=a+h$
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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  Hint
  To draw the secant line online, drag the $n_s$ slider (for number of secants) to 1, and drag the blue points to move the blue line to match the required secant line.
  There aren't any shortcuts to calculating the slope of the secant line. Just calculate the two points on the secant line and find the 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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  Hint
  To sketch the first tangent, drag $n_t$ (for number of tangents) to 1, and move the maroon point to the proper location to specify the required tangent line.
  (You won't get full credit for the tangents and number of tangents on the applet until you also draw the second tangent line described below.)
  
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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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  Hint
  To sketch the second tangent, drag $n_t$ (for number of tangents) to 2, and move the orange point to the proper location to specify the required tangent line.
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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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  Hint
  For the equation of the line, note that $x$ is on the vertical axis (playing the role $y$ usually does) and $t$ is on the horizontal axis (playing the role that $x$ usually does). So an equation for a line could be written in the form $x=mt+b$ with slope $m$ and vertical intercept $b$. (Vertical intercept is what you'd usually call $y$-intercept, but here's it's not $y$ on the vertical axis.)
  To sketch the secant line online, drag $n_s$ to 1 and drag the blue points to specify the secant line. If you inadvertently move a point off the graph, changing $n_s$ will move it back toward the center so you can reach it again.
  
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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?
  $x=$
  
  Need help?Hide help
  
  Hint
  If, on average, she was moving backward on the branch, her velocity should be negative.
  For the equation of the line, note that $x$ is on the vertical axis (playing the role $y$ usually does) and $t$ is on the horizontal axis (playing the role that $x$ usually does). So an equation for a line could be written in the form $x=mt+b$ with slope $m$ and vertical intercept $b$. (Vertical intercept is what you'd usually call $y$-intercept, but here's it's not $y$ on the vertical axis.)
  
  To sketch the secant line online, drag $n_s$ to 2 and drag the cyan points to specify the secant line. If you inadvertently move a point off the graph, changing $n_s$ will move it back toward the center so you can reach it again.
  
  Hide help
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 $0.001$.
  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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  Hint
  Yes, you have to enter a few number into the longish formula $g(t)=4 t^{5} - 52 t^{4} + 236 t^{3} - 424 t^{2} + 240 t$ and calculate the answer. It's annoying, but not the worst thing in the world.
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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.
  
  Need help?Hide help
  
  Hint
  For the equation of the line, note that $x$ is on the vertical axis (playing the role $y$ usually does) and $t$ is on the horizontal axis (playing the role that $x$ usually does). So an equation for a line could be written in the form $x=mt+b$ with slope $m$ and vertical intercept $b$. (Vertical intercept is what you'd usually call $y$-intercept, but here's it's not $y$ on the vertical axis.)
  To sketch the tangent line online, drag $n_t$ to 1 and drag the maroon points to specify the secant line. If you inadvertently move a point off the graph, changing $n_t$ will move it back toward the center so you can reach it again.
  
  Hide help
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 1241, Fall 2020

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Math 1241, Fall 2020