Chapter 2: Space Curves
Section 2.3: Tangent Vectors
Example 2.3.9
Given the two plane curves fx=x2,gx=8−x42,x⁢≥0,
At x=1, obtain the equation of the line tangent to y=f⁡x.
Find the coordinates of the intersection of y=g⁡x and the tangent line found in Part (a).
Construct a vector from ⁡1,f1 to the point found in Part (b).
Obtain R′⁡1, the natural tangent vector at ⁡1,f1.
Show that the vectors in Parts (c) and (d) are parallel. (Hint: Show their components are proportional.)
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=dfdxx=a|f(x)x=1 x−1+f1=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=gx for the two points −36,−73 and 4,7. Because of the restriction x≥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,f1 to 4,7 is
V=47−11=36=312
If R=xx2 is the radius-vector form of the curve defined by the graph of f, then R′1=12. 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
fx=x2→assign as functionf
gx=8−x/42→assign as functiong
Part (a)
Implement the point-slope form of the tangent line and press the Enter key.
y=f′1 x−1+f1
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=gx
y=2⁢x−1,y=8−116⁢x2
→solve
x=−36,y=−73,x=4,y=7
Part (c)
Write V=…
Context Panel: Assign Name
V=47−11→assign
Write V
Context Panel: Evaluate and Display Inline
V =
Part (d)
R=x,fx→assign
Expression palette: Evaluation template
Calculus palette: Differentiation operator
Context Panel: Assign to a Name≻dR1
ⅆⅆ x Rx=a|f(x)x=1 = →assign to a namedR1
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 = →plot arrow
Write the sequence of three functions shown to the right. Press the Enter key.
Context Panel: Plots≻Plot Builder Set 0≤x≤5 Options: Range from 0 to 8 Constrained Scaling
Copy and paste the tangent vector.
fx,gx,2 x−1
x2,8−116⁢x2,2⁢x−1
→
Maple Solution - Coded
Define the function f.
f≔x→x2:
Define the function g.
g≔x→8−x/42:
Obtain, at x=1, the equation of the line tangent to the graph of f.
y=Df1⋅x−1+1
Obtain the intersection of the graph of g and the line tangent to the graph of f.
solvey=2 x−1,y=gx
Obtain the vector from the point 1,f1 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,f1
Use the notation x,fx 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≔evalmapdiff,x,fx,x,x=1
Use the Equal command from the LinearAlgebra package to show that V=3 R′.
LinearAlgebra:-Equal3⋅dR,V
true
p1≔plotfx,gx,2 x−1,x=0..5,y=0..8,color=red,blue,green:T≔VectorCalculus:-RootedVectorroot=1,f1,dR:p2≔VectorCalculus:-PlotVectorT:plots:-displayp1,p2,scaling=constrained
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