2.11: Homework- Instantaneous Velocity

Last updated
Save as PDF

Page ID: 88635

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

The position of a falling object follows the equation \(f(t) = -16 t^2 + 64\) from \(t = 0\) to \(t = 2\).
1. Verify that the points \((1, 48)\) and \((2, 0)\) are on the curve by computing \(f(1)\) and \(f(2)\) and verifying you get \(48\) and \(0\).
  This seems to work.
  ans
2. Compute the slope of the line going through \((1, 48)\) and \((2, 0)\).
  -48
  ans
3. Verify that the points \((1, 48)\) and \((1 + h, -16 h^2 - 32 h + 48)\) lie on the curve. You’ve already done \((1, 48)\), so now just simplify \(f(1 + h)\) and verify you get \(-16 h^2 - 32 h + 48\).
  This seems to work.
  ans
4. Compute the slope of the line through \((1, 48)\) and \((1 + h, -16 h^2 - 32 h + 48)\) (hint: you should get \(-16h - 32\)!)
The position of a falling object follows the equation \(f(t) = -5 t^2 + 45\) from \(t = 0\) to \(t = 3\).
1. Verify that the points \((2, 25)\) and \((3, 0)\) line on the curve, and compute the slope through these two points.
  The slope is \(-25\)
  ans
2. Verify that the points \((2, 25)\) and \((2 + h, -5 h^2 - 20h + 25)\) lie on the curve, and compute the slope of the line through these two points.
The slope is \(-5h - 20\).
ans
Let \(g(t) = -10t^2 + 2000\) represent a population of goats, where \(g(t)\) is measured in goats and \(t\) is measured in years. Suppose \(t\) only works on the range from \(0\) to \(10\). This population is stabilizing during this period.
1. Sketch a graph of \(g(t)\).
2. Find the slope of the secant line hitting \(g(t)\) at \(t = 2\) and \(t = 3\).
  \(-50\) goats per year
  ans
3. The slope of the secant line hitting \(g(t)\) at \(t = 2\) and \(t = 2+h\).
  \(-40-10h\) goats per year
  ans
4. What is \(\lim_{h \to 0}\) for your answer in part (c)?
  \(-40\) goats per year.
  ans
5. How quickly is the goat population growing at \(t = 2\)?
  \(-40\) goats per year.
  ans