Example 2-3-9 - Maple Help

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Chapter 2: Space Curves

Section 2.3: Tangent Vectors

Example 2.3.9

Given the two plane curves $f (x) = x^{2}, g (x) = 8 - {(\frac{x}{4})}^{2}, x \geq 0$ ,

a)	At $x = 1$ , obtain the equation of the line tangent to $y = f (x)$ .

b)	Find the coordinates of the intersection of $y = g (x)$ and the tangent line found in Part (a).

c)	Construct a vector from $(1, f (1))$ to the point found in Part (b).

d)	Obtain $R' (1)$ , the natural tangent vector at $(1, f (1))$ .

e)	Show that the vectors in Parts (c) and (d) are parallel. (Hint: Show their components are proportional.)

f)	Draw both curves, the tangent line (Part (a)), and the tangent vector (Part (d)).

Solution

Mathematical Solution

•	At $x = 1$ , the line tangent to the graph of $f$ is given by

$y = (\binom{\frac{df}{dx}}{} | \binom{}{x = 1}) (x - 1) + f (1) = 2 x - 1$

•	The intersection of this tangent line with the graph of $g$ is obtained by solving the equations $y = 2 x - 1$ and $y = g (x)$ for the two points $(- 36, - 73)$ and $(4, 7)$ . Because of the restriction $x \geq 0$ , only the second solution is accepted.

>	use plots, VectorCalculus in module() local p1,p2,p3,T; T:=RootedVector(root=[1,1],<1,2>); p1:=PlotVector(T); p2:=plot([x^2,8-(x/4)^2,2*x-1],x=0..5,y=0..8,color=[red,blue,green]); p3:=display(p1,p2,scaling=constrained); print(p3); end module: end use:

Figure 2.3.9(a) Graph of $f, g, T$ and the tangent line (green)

•	The vector from $(1, f (1))$ to $(4, 7)$ is

$V = [\begin{array}{c} 4 \\ 7 \end{array}] - [\begin{array}{c} 1 \\ 1 \end{array}] = [\begin{array}{c} 3 \\ 6 \end{array}] = 3 [\begin{array}{c} 1 \\ 2 \end{array}]$

•	If $R = [\begin{array}{c} x \\ x^{2} \end{array}]$ is the radius-vector form of the curve defined by the graph of $f$ , then $R' (1) = [\begin{array}{c} 1 \\ 2 \end{array}]$ . Clearly, this tangent vector has the same direction as $V$ since these two vectors are proportional.

•	Figure 2.3.9(a) shows the graph of $f$ in red, the graph of $g$ in blue, and of the tangent line in green. The vector $R' (1)$ is represented by the black arrow.

Maple Solution - Interactive

Initialize

•	Tools≻Load Package: Student Multivariate Calculus

Loading Student:-MultivariateCalculus

•	Context Panel: Assign Function

$f (x) = x^{2}$ $\overset{assign as function}{\to}$ $f$

•	Context Panel: Assign Function

$g (x) = 8 - {(x / 4)}^{2}$ $\overset{assign as function}{\to}$ $g$

Part (a)

•	Implement the point-slope form of the tangent line and press the Enter key.

$y = f' (1) (x - 1) + f (1)$

$y = 2 x - 1$

Part (b)

•	Write the sequence of equations to be solved and press the Enter key.

•	Context Panel: Solve≻Solve

$y = 2 x - 1, y = g (x)$

$y = 2 x - 1, y = 8 - \frac{1}{16} x^{2}$

$\overset{solve}{\to}$

$\{x = - 36, y = - 73\}, \{x = 4, y = 7\}$

Part (c)

•	Write $V = \dots$

•	Context Panel: Assign Name

$V = [\begin{array}{c} 4 \\ 7 \end{array}] - [\begin{array}{c} 1 \\ 1 \end{array}]$ $\overset{assign}{\to}$

•

Write $V$

•	Context Panel: Evaluate and Display Inline

$V$ =

Part (d)

•	Context Panel: Assign Name

$R = ⟨x, f (x)⟩$ $\overset{assign}{\to}$

•	Expression palette: Evaluation template

•	Calculus palette: Differentiation operator

•	Context Panel: Assign to a Name≻dR1

$\binom{\frac{&DifferentialD;}{&DifferentialD; x} R}{} | \binom{}{x = 1}$ = $\overset{assign to a name}{\to}$ $dR1$

Part (e)

•	Write V and 3 dR1, and press the Enter key.

•	Since these two vectors are obviously equal, V and dR1 are proportional.

$V, 3 dR1$

Part (f)

•	Write dR1.

•	Context Panel: Evaluate and Display inline

•	Context Panel: Plots: Arrow from point $x = 1, y = 1$

$dR1$ = $\overset{plot arrow}{\to}$

•	Write the sequence of three functions shown to the right. Press the Enter key.

•	Context Panel: Plots≻Plot Builder Set $0 \leq x \leq 5$ Options: Range from 0 to 8 Constrained Scaling

•	Copy and paste the tangent vector.

$f (x), g (x), 2 x - 1$

$x^{2}, 8 - \frac{1}{16} x^{2}, 2 x - 1$

$\to$

Maple Solution - Coded

Initialize

•	Define the function $f$ .

$f ≔ x \to x^{2} &colon;$

•	Define the function $g$ .

$g ≔ x \to 8 - {(x / 4)}^{2} &colon;$

Part (a)

•	Obtain, at $x = 1$ , the equation of the line tangent to the graph of $f$ .

$y = D (f) (1) \cdot (x - 1) + 1$

$y = 2 x - 1$

Part (b)

•	Obtain the intersection of the graph of $g$ and the line tangent to the graph of $f$ .

$solve (\{y = 2 x - 1, y = g (x)\})$

$\{x = - 36, y = - 73\}, \{x = 4, y = 7\}$

Part (c)

•	Obtain the vector from the point $(1, f (1))$ to $(4, 7)$ , the point of intersection of the line tangent to the graph of $f$ and the graph of $g$ .

$V ≔ ⟨4, 7⟩ - ⟨1, f (1)⟩$

Part (d)

•	Use the notation $⟨x, f (x)⟩$ for the position-vector representation of the graph of $f$ . Then use the map command to apply the diff command to each component of the vector. Finally, use the eval command to evaluate the tangent vector at $x = 1$ .

$dR ≔ eval (map (diff, ⟨x, f (x)⟩, x), x = 1)$

Part (e)

•	Use the Equal command from the LinearAlgebra package to show that $V = 3 R'$ .

$LinearAlgebra :- Equal (3 \cdot dR, V)$

$true$

Part (f)

$p_{1} ≔ plot ([f (x), g (x), 2 x - 1], x = 0 .. 5, y = 0 .. 8, color = [red, blue, green]) &colon; T ≔ VectorCalculus :- RootedVector (root = [1, f (1)], dR) &colon; p_{2} ≔ VectorCalculus :- PlotVector (T) &colon; plots :- display (p_{1}, p_{2}, scaling = constrained)$

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