TBIL-Cal Parametric/Vector Derivatives (CO2)

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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 = 2 t - 1

and

y = (2 t - 1) (2 t - 5) .

The coordinate on this graph at

t = 2

is

(3, - 3) .

(a)

Which of the following equations of

x, y

describes the graph of these paramteric equations?

$y = 2 x (x + 2) = 2 x^{2} + 2 x$
$y = 2 x (x - 2) = 2 x^{2} - 2 x$
$y = x (x + 4) = x^{2} + 4 x$
$y = x (x - 4) = x^{2} - 4 x$

(b)

Which of the following describes the slope of the line tangent to the graph at the point

(3, - 3) ?

$\frac{d y}{d x} = 2 x + 4,$ which is $10$ when $x = 3 .$
$\frac{d y}{d x} = 2 x + 4,$ which is $8$ when $t = 2 .$
$\frac{d y}{d x} = 2 x - 4,$ which is $2$ when $x = 3 .$
$\frac{d y}{d x} = 2 x - 4,$ which is $0$ when $t = 2 .$

(c)

Note that the parametric equation for

y

simplifies to

y = 4 t^{2} - 12 t + 5 .

What do we get for the derivatives

\frac{d x}{d t}

of

x = 2 t - 1

and

\frac{d y}{d t}

for

y = 4 t^{2} - 12 t + 5 ?

$\frac{d x}{d t} = 2$ and $\frac{d y}{d t} = 8 t - 12 .$
$\frac{d x}{d t} = - 1$ and $\frac{d y}{d t} = 8 t - 12 .$
$\frac{d x}{d t} = 2$ and $\frac{d y}{d t} = 6 t + 5 .$
$\frac{d x}{d t} = - 1$ and $\frac{d y}{d t} = 6 t + 5 .$

(d)

It follows that when

t = 2,

\frac{d x}{d t} = 2

and

\frac{d y}{d t} = 4 .

Which of the following conjectures seems most likely?

The slope $\frac{d y}{d x}$ could also be found by computing $\frac{d x}{d t} + \frac{d y}{d t} .$
The slope $\frac{d y}{d x}$ could also be found by computing $\frac{d y / d t}{d x / d t} .$
The slope $\frac{d y}{d x}$ is always equal to $\frac{d x}{d t} .$
The slope $\frac{d y}{d x}$ is always equal to $\frac{d y}{d t} .$

Fact 7.2.2.

Suppose

x

is a function of

t,

and

y

may be thought of as a function of either

x

or

t .

Then the Chain Rule requires that

\frac{d y}{d t} = \frac{d y}{d x} \frac{d x}{d t} .

This provides the slope formula for parametric equations:

\frac{d y}{d x} = \frac{d y / d t}{d x / d t} .

Activity 7.2.3.

Let’s draw the picture of the line tangent to the parametric equations

x = 2 t - 1

and

y = (2 t - 1) (2 t - 5)

when

t = 2 .

(a)

Use a

t, x, y

chart to sketch the parabola given by these parametric equations for

0 \leq t \leq 3,

including the point

(3, - 3)

when

t = 2 .

(b)

Earlier we determined that the slope of the tangent line was

2 .

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.

$y = 2 x - 10$
$y = 2 x - 9$
$y = 2 x - 8$
$y = 2 x - 7$

Activity 7.2.4.

Consider the vector equation

\vec{r} (t) = ⟨ 3 t^{2} - 9, t^{3} - 3 t ⟩ .

(a)

What are the corresponding parametric equations and their derivatives?

$y = 3 t^{2} - 9$ and $x = t^{3} - 3 t;$ $\frac{d y}{d t} = 9 t$ and $\frac{d x}{d t} = 3 t - 6$
$x = 3 t^{2} - 9$ and $y = t^{3} - 3 t;$ $\frac{d x}{d t} = 9 t$ and $\frac{d y}{d t} = 3 t - 6$
$y = 3 t^{2} - 9$ and $x = t^{3} - 3 t;$ $\frac{d y}{d t} = 6 t$ and $\frac{d x}{d t} = 3 t^{2} - 3$
$x = 3 t^{2} - 9$ and $y = t^{3} - 3 t;$ $\frac{d x}{d t} = 6 t$ and $\frac{d y}{d t} = 3 t^{2} - 3$

(b)

The formula

\frac{d y}{d x} = \frac{d y / d t}{d x / d t}

allows us to compute slopes as which of the following functions of

t ?

$\frac{6 t}{t^{2} + 3}$
$\frac{6 t}{t^{2} + 1}$
$\frac{t^{2} - 1}{2 t}$
$\frac{2 t}{3 t^{2} - 1}$

(c)

Find the point, tangent slope, and tangent line equation (recall

y - y_{0} = m (x - x_{0})

) corresponding to the parameter

t = - 3 .

Point $(- 12, 9),$ slope $- \frac{4}{3},$ EQ $y = - \frac{4}{3} x - 7$
Point $(18, - 18),$ slope $- \frac{4}{3},$ EQ $y = - \frac{4}{3} x + 6$
Point $(- 12, 9),$ slope $\frac{3}{4},$ EQ $y = \frac{3}{4} x - 8$
Point $(18, - 18),$ slope $\frac{3}{4},$ EQ $y = \frac{3}{4} x + 5$

Subsection 7.2.2 Videos

Figure 171. Video for CO2

Subsection 7.2.3 Exercises

Exercises available at https://tbil.org/calculus/preview/exercises/#/bank/CO2/