MAT2615 Assignment 2 (COMPLETE ANSWERS) 2026 - DUE June 2026;.
Consider the R2 − R function f defined by f (x, y) = 1 − x2 − y2. Let C be
the contour curve of f through the point (1,−1), let L be the tangent to C at
(x, y) = (1, 1) and let V be the tangent plane to f at (x, y) = (1, 1). (a) Find
the equation of the curve C. (2) (b) Find a vector in R2 that is perpendicular
to C at (x, y) = (1, 1). (2) (c) Find the Cartesian equation of the line L. (3) (d)
Find a vector in R3 that is perpendicular to the graph of f at the point (x, y, z)
= (1, 1, 3).(3) (e) Find the Cartesian equation of the plane V. (3) (f) Draw a
sketch to visualize the graph of f , together with appropriate sections of the
line L and the plane V. Also show the vectors that you obtained in (b) and (d)
on your sketch. (3) Hints • Study Definitions 3.2.5 and 3.2.9. Note that the
level of C is given by f (1, 1). • By a vector perpendicular to a curve at a
given point, we mean a vector perpendicular to the tangent to the curve at
that point. Use Theorem 7.9.1 to find a vector perpendicular to C at the point
(1, 1). • Study Remark 2.12.2(1) and use Definition 2.12.1 to find the
Cartesian equation of L. Or, equivalently, use Definition 7.9.6. (Note that, in
the case n = 2, the formula in Definition 7.9.6 gives a Cartesian equation for
a tangent to a contour curve.) • By a vector perpendicular to a surface at a
given point, we mean a vector perpendicular to the tangent plane to the
surface at that point. Define an R3 − R function g such that the graph of f is
a contour surface of g, and then use Theorem 7.9.3 to find a vector
perpendicular to V at the point (1, 1, 3). • Use Definition 2.12.1 or Definition
7.9.6 (with g in the place of f ) to find the equation of V, or use Definition
7.5.3. (Read Remark 7.5.4(2).) [16] 2.
(Chapter 9) Consider the R2 − R function f defined by f (x, y) = sin x cos y.
(a) Find the second order Taylor Polynomial of f about the point π 4 , π 4 .
Leave your answer in the form of a polynomial in x − π 4 and y − π 4 .
(This form is convenient for evaluating function values at points near π 4 , π
4 .) (6) (b) Use your answer to (a) to estimate the value of e0,1 ln 0, 9.
Compare your answer with the approximation given by a pocket calculator.
(2) [8] 2 Downloaded by Edge Tutor () lOMoARcPSD| MAT2615/AS2/0/2026 3.
(Sections 11.1 - 11.3, 7.5 and 9.3) (a) State the Implicit Function Theorem for
an equation in the three variables, x y and z.(2) (b) Use the Implicit Function
Theorem to show that the equation xyz = cos (x + y + z) has a smooth
unique local solution of the form z = g(x, y) about the point (0, 0, π2 ). Then
find a linear approximation for g about (0, 0). Hints • Use the method of
Example 11.2.6, but take into account that you are dealing with an equation
in three variables here and that g in this case is a function of two variables. •
Before you apply the Implicit Function Theorem you should show that all the
necessary conditions are satisfied. • Study Remark 11.3.3(1) and Remark
9.3.6(2). (6) [8] 4. (Sections 1.3, 11.2 and 11.3) Consider the 2-dimensional
vector field F defined by F(x, y) =
MAT2615 – APPLIED MATHEMATICS
, ASSIGNMENT 2: COMPLETE ANSWERS AND
EXPLANATIONS
STUDY YEAR: 2026
DOCUMENT INFORMATION (FOR QUALITY &
VISIBILITY)
Title
MAT2615 Assignment 2 – Complete Solutions with Detailed Explanations (2026)
Module
Applied Mathematics (Multivariable Calculus)
Description
This document contains fully worked and thoroughly explained solutions to all questions in
MAT2615 Assignment 2 (2026). The assignment covers contour curves, gradients, tangent lines
and planes, Taylor polynomials, the Implicit Function Theorem, and linear approximations. Each
solution is presented in a clear academic format suitable for university assessment and online
academic platforms.
Tags
MAT2615, Applied Mathematics, Assignment 2, Multivariable Calculus, 2026, Complete
Solutions, Taylor Polynomials, Implicit Function Theorem, Tangent Planes, Contour Curves
PAGE 1 – COVER PAGE
Module Code: MAT2615
Assignment: Assignment 2
Year: 2026
Institution: UNISA
Document Type: Fully Worked Solutions
Consider the R2 − R function f defined by f (x, y) = 1 − x2 − y2. Let C be
the contour curve of f through the point (1,−1), let L be the tangent to C at
(x, y) = (1, 1) and let V be the tangent plane to f at (x, y) = (1, 1). (a) Find
the equation of the curve C. (2) (b) Find a vector in R2 that is perpendicular
to C at (x, y) = (1, 1). (2) (c) Find the Cartesian equation of the line L. (3) (d)
Find a vector in R3 that is perpendicular to the graph of f at the point (x, y, z)
= (1, 1, 3).(3) (e) Find the Cartesian equation of the plane V. (3) (f) Draw a
sketch to visualize the graph of f , together with appropriate sections of the
line L and the plane V. Also show the vectors that you obtained in (b) and (d)
on your sketch. (3) Hints • Study Definitions 3.2.5 and 3.2.9. Note that the
level of C is given by f (1, 1). • By a vector perpendicular to a curve at a
given point, we mean a vector perpendicular to the tangent to the curve at
that point. Use Theorem 7.9.1 to find a vector perpendicular to C at the point
(1, 1). • Study Remark 2.12.2(1) and use Definition 2.12.1 to find the
Cartesian equation of L. Or, equivalently, use Definition 7.9.6. (Note that, in
the case n = 2, the formula in Definition 7.9.6 gives a Cartesian equation for
a tangent to a contour curve.) • By a vector perpendicular to a surface at a
given point, we mean a vector perpendicular to the tangent plane to the
surface at that point. Define an R3 − R function g such that the graph of f is
a contour surface of g, and then use Theorem 7.9.3 to find a vector
perpendicular to V at the point (1, 1, 3). • Use Definition 2.12.1 or Definition
7.9.6 (with g in the place of f ) to find the equation of V, or use Definition
7.5.3. (Read Remark 7.5.4(2).) [16] 2.
(Chapter 9) Consider the R2 − R function f defined by f (x, y) = sin x cos y.
(a) Find the second order Taylor Polynomial of f about the point π 4 , π 4 .
Leave your answer in the form of a polynomial in x − π 4 and y − π 4 .
(This form is convenient for evaluating function values at points near π 4 , π
4 .) (6) (b) Use your answer to (a) to estimate the value of e0,1 ln 0, 9.
Compare your answer with the approximation given by a pocket calculator.
(2) [8] 2 Downloaded by Edge Tutor () lOMoARcPSD| MAT2615/AS2/0/2026 3.
(Sections 11.1 - 11.3, 7.5 and 9.3) (a) State the Implicit Function Theorem for
an equation in the three variables, x y and z.(2) (b) Use the Implicit Function
Theorem to show that the equation xyz = cos (x + y + z) has a smooth
unique local solution of the form z = g(x, y) about the point (0, 0, π2 ). Then
find a linear approximation for g about (0, 0). Hints • Use the method of
Example 11.2.6, but take into account that you are dealing with an equation
in three variables here and that g in this case is a function of two variables. •
Before you apply the Implicit Function Theorem you should show that all the
necessary conditions are satisfied. • Study Remark 11.3.3(1) and Remark
9.3.6(2). (6) [8] 4. (Sections 1.3, 11.2 and 11.3) Consider the 2-dimensional
vector field F defined by F(x, y) =
MAT2615 – APPLIED MATHEMATICS
, ASSIGNMENT 2: COMPLETE ANSWERS AND
EXPLANATIONS
STUDY YEAR: 2026
DOCUMENT INFORMATION (FOR QUALITY &
VISIBILITY)
Title
MAT2615 Assignment 2 – Complete Solutions with Detailed Explanations (2026)
Module
Applied Mathematics (Multivariable Calculus)
Description
This document contains fully worked and thoroughly explained solutions to all questions in
MAT2615 Assignment 2 (2026). The assignment covers contour curves, gradients, tangent lines
and planes, Taylor polynomials, the Implicit Function Theorem, and linear approximations. Each
solution is presented in a clear academic format suitable for university assessment and online
academic platforms.
Tags
MAT2615, Applied Mathematics, Assignment 2, Multivariable Calculus, 2026, Complete
Solutions, Taylor Polynomials, Implicit Function Theorem, Tangent Planes, Contour Curves
PAGE 1 – COVER PAGE
Module Code: MAT2615
Assignment: Assignment 2
Year: 2026
Institution: UNISA
Document Type: Fully Worked Solutions
