MAT2615
Assignment 2
2026
, MAT2615 Assignment 2 - 2026
Calculus in Higher Dimensions
University of South Africa (UNISA)
Question 1: Contour Curves and Tangent Planes
Given: f(x, y) = 1 − x² − y², with contour curve C through point (1, −1), and we analyze at point
(1, 1).
(a) Equation of curve C [2 marks]
Solution:
First, find the level by evaluating f at (1, 1):
f(1, 1) = 1 − 1² − 1² = 1 − 1 − 1 = −1
The contour curve C is where f(x, y) = −1:
1 − x² − y² = −1
x² + y² = 2
Answer: The equation of curve C is x² + y² = 2
(b) Vector perpendicular to C at (1, 1) [2 marks]
Solution:
By Theorem 7.9.1, the gradient ∇f is perpendicular to the contour curve at any point.
Calculate the gradient:
∇f(x, y) = (∂f/∂x, ∂f/∂y) = (−2x, −2y)
At point (1, 1):
Assignment 2
2026
, MAT2615 Assignment 2 - 2026
Calculus in Higher Dimensions
University of South Africa (UNISA)
Question 1: Contour Curves and Tangent Planes
Given: f(x, y) = 1 − x² − y², with contour curve C through point (1, −1), and we analyze at point
(1, 1).
(a) Equation of curve C [2 marks]
Solution:
First, find the level by evaluating f at (1, 1):
f(1, 1) = 1 − 1² − 1² = 1 − 1 − 1 = −1
The contour curve C is where f(x, y) = −1:
1 − x² − y² = −1
x² + y² = 2
Answer: The equation of curve C is x² + y² = 2
(b) Vector perpendicular to C at (1, 1) [2 marks]
Solution:
By Theorem 7.9.1, the gradient ∇f is perpendicular to the contour curve at any point.
Calculate the gradient:
∇f(x, y) = (∂f/∂x, ∂f/∂y) = (−2x, −2y)
At point (1, 1):
