High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize University Calculus 3 Textbook > Vector Functions

Functions of 2 and 3 Variables & Contour Curves

Quadratic Surfaces and Level CurvesPrevious Section
Tangent, Normal, Binormal Vectors - Applications of Vector FunctionsNext Section
Popular Courses
Find My Course
0:00 / 0:00

Functions of 2 & 3 Variables and Contour Curves

In R2\colorOne{\mathbb{R}^2}

A function of 2 variables, such as f(x,y), f(x,y),~relates a sets of points in the (x,y)(x, y) format to a number
  • k=f(x,y); k∈Rk=f(x,y); ~k\in\mathbb{R}

In R3\colorOne{\mathbb{R}^3}

A function of 3 variables, such as f(x,y,z),f(x,y,z), relates a set of points in the (x,y,z)(x,y,z) format to a number
  • k=f(x,y,z); k∈Rk=f(x,y,z); ~k\in\mathbb{R}
  • When a function of 3 variables formats to a number, we can graph the contour/level curves, since level curves of a function z=f(x,y) z=f(x,y)~ are 2D curves we get by setting z=k, k∈Rz=k,~k\in\mathbb{R}.
Wize Tip
Contour curves can also be written as f(x, y, k) = 0 if the form of the function is in the format f(x, y, z) =0.

Example

Lets look at a cone, z=x2+y2z=\sqrt{x^2+y^2}z=kz=k, and we let , then we get a cone in 3D that, in 2D, are concentric circles with the radius, r=kr=k.



Now, cut each section of the cone at varying values of zz.

And map the cross-sectional area in 2D to obtain:

We can think of contours in terms of the intersection of the surface that is given by z = f(x, y) and the plane z = k. Therefore,

Wize Concept
The contour is the intersection of the surface and the plane

PAGE BREAK

Vector Functions

A vector function takes on one or more variables and returns a vector.
  • In R2\mathbb{R}^{2}, the vector function is r(t)⃗= <f(t), g(t)>\vec{r(t)}=~<f(t),~g(t)>
  • In R3, \mathbb{R}^{3},~the vector function is r(t)⃗= <f(t), g(t), h(t)>\vec{r(t)}=~<f(t),~g(t),~h(t)>
where f(t), g(t), h(t) f(t),~g(t),~h(t)~are the component functions of the vector function.

Determining the domain of a vector function requires one to set the domain on each component.

Example

The vector <cos⁡t,ln⁡(4−t),t+1>\Big<\cos{t},\ln(4-t),\sqrt{t+1}\Big> has a domain of [−1, 4)[-1,~4) since the domain for each component is:

cos⁡tln⁡(4−t)t+1−1≤cos⁡t≤14−t>0t+1≥0Domain‾: [−1,1]Domain‾: t<4Domain‾: t≥−1\begin{array}{c||c||c} \colorSix{\cos{t}}&\colorSix{\ln(4-t)}&\colorSix{\sqrt{t+1}}\\\hline \\ -1\leq{}\cos{t}\leq1&4-t>0&t+1\geq{0}\\\\\hline\\ \text{\underline{Domain}:}~\colorThree{[-1,1]}&\text{\underline{Domain}:}~\colorThree{t<4}&\text{\underline{Domain}:}~\colorThree{t\geq{-1}} \end{array}

0:00 / 0:00

Example

Evaluate the function f(x,y)=3−x23−y227f(x,y)=\sqrt{3-\frac{x^2}{3}-\frac{y^2}{27}} at the point (1,1)(1,1). Then, determine and sketch the domain of this function.

Evaluating f(1,1)\colorOne{f(1,1)}:


f(1,1)=3−1/3−1/27=7127f(1,1)=\sqrt{3-1/3-1/27}=\sqrt{\frac{71}{27}}

The domain is:


x23+y227=3Divide by 3x29+y281=1Ellipse centered at (0,0) andwith Minor/Major axis of 3 & 9 respectively\begin{array}{rclll} \dfrac{x^2}{3}+\dfrac{y^2}{27}&=&3&&\color{magenta}{\text{\scriptsize{Divide by 3}}}\\\\ \dfrac{x^2}{9}+\dfrac{y^2}{81}&=&1&&\color{magenta}{\text{\scriptsize{Ellipse centered at }(0,0)~\text{\scriptsize{and}}}}\\ &&&&\color{magenta}{\text{\scriptsize{with Minor/Major axis of 3 }\&~9~\text{\scriptsize{respectively}}}} \end{array}

−3≤x≤3;   −9≤y≤9-3\leq{x}\leq3;~~~-9\leq{y}\leq{9}













Since we know that:

3−x23−y227≥0x23+y227≤3x29+y281≤1\begin{array}{rcl} 3-\dfrac{x^2}{3}-\dfrac{y^2}{27}&\ge&0\\\\ \dfrac{x^2}{3}+\dfrac{y^2}{27}&\le&3\\\\\dfrac{x^2}{9}+\dfrac{y^2}{81}&\le&1 \end{array}
This is an ellipse in 2D.
0:00 / 0:00

Example

Sketch the graph of r(t)⃗=<9cos⁡t, 9sin⁡t, 5>\vec{r(t)}=<9\cos{t},~9\sin{t},~5>.

If we notice the x and y component functions, we see that it is nothing more than a circle with radius 9
cos⁡2θ+sin⁡2θ=r2Equation of a Circle(9cos⁡t)2+(9sin⁡t)2=r281cos⁡2t+81sin⁡2t=r281(cos⁡2t+sin⁡2t)=r2Pythagorean  Identity∴ r=9\begin{array}{rclll} \cos^{2}\theta+\sin^{2}\theta&=&r^{2}&&{\color{magenta}{\text{\scriptsize{Equation~of~a~Circle}}}}\\\\ (9\cos{t})^{2}+(9\sin{t})^{2}&=&r^2\\\\ 81\cos^{2}t+81\sin^{2}t&=&r^2\\\\ 81\big({\color{magenta}{\cos^{2}t+\sin^2{t}}}\big)&=&r^{2}&&{\color{magenta}{\text{\scriptsize{Pythagorean ~Identity}}}}\\\\ \therefore~r&=&9 \end{array}
If we look at the z component function, it indicates that our circle will be level with z = 5.

checklist
Mark Yourself Question
  1. Grab a piece of paper and try this problem yourself.
  2. When you're done, check the "I have answered this question" box below.
  3. View the solution and report whether you got it right or wrong.

Practice

Sketch the contour curves for 6z+3y2−x=0.6z+3y^{2}-x=0.