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Section 7.2 Parametric/Vector Derivatives (CO2)
Learning Outcomes
Compute derivatives and tangents related to two-dimensional parametric/vector equations.
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 parametric equations?
\(\displaystyle y=2x(x+2)=2x^2+2x\)
\(\displaystyle y=2x(x-2)=2x^2-2x\)
\(\displaystyle y=x(x+4)=x^2+4x\)
\(\displaystyle y=x(x-4)=x^2-4x\)
(b) Which of the following describes the slope of the line tangent to the graph at the point
\((3,-3)\text{?}\)
\(\frac{dy}{dx}=2x+4\text{,}\) which is
\(10\) when
\(x=3\text{.}\)
\(\frac{dy}{dx}=2x+4\text{,}\) which is
\(8\) when
\(t=2\text{.}\)
\(\frac{dy}{dx}=2x-4\text{,}\) which is
\(2\) when
\(x=3\text{.}\)
\(\frac{dy}{dx}=2x-4\text{,}\) 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
\(y=4t^2-12t+5\text{?}\)
\(\frac{dx}{dt}=2\) and
\(\frac{dy}{dt}=8t-12\text{.}\)
\(\frac{dx}{dt}=-1\) and
\(\frac{dy}{dt}=8t-12\text{.}\)
\(\frac{dx}{dt}=2\) and
\(\frac{dy}{dt}=6t+5\text{.}\)
\(\frac{dx}{dt}=-1\) 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?
The slope
\(\frac{dy}{dx}\) could also be found by computing
\(\frac{dx}{dt}+\frac{dy}{dt}\text{.}\)
The slope
\(\frac{dy}{dx}\) could also be found by computing
\(\frac{dy/dt}{dx/dt}\text{.}\)
The slope
\(\frac{dy}{dx}\) is always equal to
\(\frac{dx}{dt}\text{.}\)
The slope
\(\frac{dy}{dx}\) is always equal to
\(\frac{dy}{dt}\text{.}\)
Fact 7.2.2 .
Suppose \(x\) is a function of \(t\text{,}\) and \(y\) may be thought of as a function of either \(x\) or \(t\text{.}\) Then the Chain Rule requires that
\begin{equation*}
\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}\text{.}
\end{equation*}
This provides the slope formula for parametric equations:
\begin{equation*}
\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\text{.}
\end{equation*}
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.
\(\displaystyle y=2x-10\)
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?
\(y=3t^2-9\) and
\(x=t^3-3t\text{;}\) \(\frac{dy}{dt}=9t\) and
\(\frac{dx}{dt}=3t-6\)
\(x=3t^2-9\) and
\(y=t^3-3t\text{;}\) \(\frac{dx}{dt}=9t\) and
\(\frac{dy}{dt}=3t-6\)
\(y=3t^2-9\) and
\(x=t^3-3t\text{;}\) \(\frac{dy}{dt}=6t\) and
\(\frac{dx}{dt}=3t^2-3\)
\(x=3t^2-9\) 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
\(t\text{?}\)
\(\displaystyle \frac{6t}{t^2+3}\)
\(\displaystyle \frac{6t}{t^2+1}\)
\(\displaystyle \frac{t^2-1}{2t}\)
\(\displaystyle \frac{2t}{3t^2-1}\)
(c) Find the point, tangent slope, and tangent line equation (recall
\(y-y_0=m(x-x_0)\) ) corresponding to the parameter
\(t=-3\text{.}\)
Point
\((-12,9)\text{,}\) slope
\(-\frac{4}{3}\text{,}\) EQ
\(y=-\frac{4}{3}x-7\)
Point
\((18,-18)\text{,}\) slope
\(-\frac{4}{3}\text{,}\) EQ
\(y=-\frac{4}{3}x+6\)
Point
\((-12,9)\text{,}\) slope
\(\frac{3}{4}\text{,}\) EQ
\(y=\frac{3}{4}x-8\)
Point
\((18,-18)\text{,}\) slope
\(\frac{3}{4}\text{,}\) EQ
\(y=\frac{3}{4}x+5\)
Subsection 7.2.2 Videos
Figure 166. Video for CO2
Subsection 7.2.3 Exercises
