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Calculus for Team-Based Inquiry Learning: PREVIEW Edition — Instructor Version

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Section 7.2 Parametric/Vector Derivatives (CO2)

Subsection 7.2.1 Activities

Activity 7.2.1.

Consider the parametric equations \(x=2t-1\) and \(y=(2t-1)(2t-5)\text{.}\) The coordinate on this graph at \(t=2\) is \((3,-3)\text{.}\)
(a)
Which of the following equations of \(x,y\) describes the graph of these paramteric equations?
  1. \(\displaystyle y=2x(x+2)=2x^2+2x\)
  2. \(\displaystyle y=2x(x-2)=2x^2-2x\)
  3. \(\displaystyle y=x(x+4)=x^2+4x\)
  4. \(\displaystyle y=x(x-4)=x^2-4x\)
(b)
Which of the following describes the slope of the line tangent to the graph at the point \((3,-3)\text{?}\)
  1. \(\frac{dy}{dx}=2x+4\text{,}\) which is \(10\) when \(x=3\text{.}\)
  2. \(\frac{dy}{dx}=2x+4\text{,}\) which is \(8\) when \(t=2\text{.}\)
  3. \(\frac{dy}{dx}=2x-4\text{,}\) which is \(2\) when \(x=3\text{.}\)
  4. \(\frac{dy}{dx}=2x-4\text{,}\) which is \(0\) when \(t=2\text{.}\)
(c)
Note that the parametric equation for \(y\) simplifies to \(y=4t^2-12t+5\text{.}\) What do we get for the derivatives \(\frac{dx}{dt}\) of \(x=2t-1\) and \(\frac{dy}{dt}\) for \(y=4t^2-12t+5\text{?}\)
  1. \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=8t-12\text{.}\)
  2. \(\frac{dx}{dt}=-1\) and \(\frac{dy}{dt}=8t-12\text{.}\)
  3. \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=6t+5\text{.}\)
  4. \(\frac{dx}{dt}=-1\) and \(\frac{dy}{dt}=6t+5\text{.}\)
(d)
It follows that when \(t=2\text{,}\) \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=4\text{.}\) Which of the following conjectures seems most likely?
  1. The slope \(\frac{dy}{dx}\) could also be found by computing \(\frac{dx}{dt}+\frac{dy}{dt}\text{.}\)
  2. The slope \(\frac{dy}{dx}\) could also be found by computing \(\frac{dy/dt}{dx/dt}\text{.}\)
  3. The slope \(\frac{dy}{dx}\) is always equal to \(\frac{dx}{dt}\text{.}\)
  4. The slope \(\frac{dy}{dx}\) is always equal to \(\frac{dy}{dt}\text{.}\)

Activity 7.2.3.

Let’s draw the picture of the line tangent to the parametric equations \(x=2t-1\) and \(y=(2t-1)(2t-5)\) when \(t=2\text{.}\)
(a)
Use a \(t,x,y\) chart to sketch the parabola given by these parametric equations for \(0\leq t\leq 3\text{,}\) including the point \((3,-3)\) when \(t=2\text{.}\)
(b)
Earlier we determined that the slope of the tangent line was \(2\text{.}\) Draw a line with slope \(2\) passing through \((3,-3)\) and confirm that it appears to be tangent.
(c)
Use the point-slope formula \(y-y_0=m(x-x_0)\) along with the slope \(2\) and point \((3,-3)\) to find the exact equation for this tangent line.
  1. \(\displaystyle y=2x-10\)
  2. \(\displaystyle y=2x-9\)
  3. \(\displaystyle y=2x-8\)
  4. \(\displaystyle y=2x-7\)

Activity 7.2.4.

Consider the vector equation \(\vec{r}(t)=\tuple{3t^2-9,t^3-3t}\text{.}\)
(a)
What are the corresponding parametric equations and their derivatives?
  1. \(y=3t^2-9\) and \(x=t^3-3t\text{;}\) \(\frac{dy}{dt}=9t\) and \(\frac{dx}{dt}=3t-6\)
  2. \(x=3t^2-9\) and \(y=t^3-3t\text{;}\) \(\frac{dx}{dt}=9t\) and \(\frac{dy}{dt}=3t-6\)
  3. \(y=3t^2-9\) and \(x=t^3-3t\text{;}\) \(\frac{dy}{dt}=6t\) and \(\frac{dx}{dt}=3t^2-3\)
  4. \(x=3t^2-9\) and \(y=t^3-3t\text{;}\) \(\frac{dx}{dt}=6t\) and \(\frac{dy}{dt}=3t^2-3\)
(b)
The formula \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\) allows us to compute slopes as which of the following functions of \(t\text{?}\)
  1. \(\displaystyle \frac{6t}{t^2+3}\)
  2. \(\displaystyle \frac{6t}{t^2+1}\)
  3. \(\displaystyle \frac{t^2-1}{2t}\)
  4. \(\displaystyle \frac{2t}{3t^2-1}\)
(c)
Find the point, tangent slope, and tangent line equation (recall \(y-y_0=m(x-x_0)\)) corresponding to the parameter \(t=-3\text{.}\)
  1. Point \((-12,9)\text{,}\) slope \(-\frac{4}{3}\text{,}\) EQ \(y=-\frac{4}{3}x-7\)
  2. Point \((18,-18)\text{,}\) slope \(-\frac{4}{3}\text{,}\) EQ \(y=-\frac{4}{3}x+6\)
  3. Point \((-12,9)\text{,}\) slope \(\frac{3}{4}\text{,}\) EQ \(y=\frac{3}{4}x-8\)
  4. Point \((18,-18)\text{,}\) slope \(\frac{3}{4}\text{,}\) EQ \(y=\frac{3}{4}x+5\)

Subsection 7.2.2 Videos

Figure 171. Video for CO2

Subsection 7.2.3 Exercises

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