MAT2615 Assignment 2 solutions 2026
UNISA
ALL ANSWERS ARE VERIFIED AND ANSWERED BEST CLEARLY, STEP BY
STEP ALL CALCULATIONS ARE SHOWN TOO
Dear Student , type or hand write this is allowed and
I’m your private tutor iQ Level
CALCULUS IN HIGHER DIMENSIONS
MAT2615
Year module
Department of Mathematical Sciences
IMPORTANT INFORMATION:
Please activate your myUnisa and myLife e-mail account and
make sure that you have regular access to the myUnisa module
website MAT2615-26-Y, as well as your group website.
Note: This is a fully online module. It is therefore, only available on
myUnisa.
, QUESTION 1
Given:
𝑓(𝑥, 𝑦) = 1 − 𝑥 2 − 𝑦 2
(a) Equation of contour curve 𝑪
A contour curve is defined by:
𝑓(𝑥, 𝑦) = constant
Level is:
𝑓(1,1) = 1 − 12 − 12 = 1 − 1 − 1 = −1
So:
1 − 𝑥 2 − 𝑦 2 = −1
Rearrange:
𝑥2 + 𝑦2 = 2
Answer:
𝐶: 𝑥 2 + 𝑦 2 = 2
(b) Vector perpendicular to 𝑪
Perpendicular vector = gradient:
∇𝑓 = (−2𝑥, −2𝑦)
At (1,1):
∇𝑓(1,1) = (−2, −2)
Answer:
(−2, −2)
UNISA
ALL ANSWERS ARE VERIFIED AND ANSWERED BEST CLEARLY, STEP BY
STEP ALL CALCULATIONS ARE SHOWN TOO
Dear Student , type or hand write this is allowed and
I’m your private tutor iQ Level
CALCULUS IN HIGHER DIMENSIONS
MAT2615
Year module
Department of Mathematical Sciences
IMPORTANT INFORMATION:
Please activate your myUnisa and myLife e-mail account and
make sure that you have regular access to the myUnisa module
website MAT2615-26-Y, as well as your group website.
Note: This is a fully online module. It is therefore, only available on
myUnisa.
, QUESTION 1
Given:
𝑓(𝑥, 𝑦) = 1 − 𝑥 2 − 𝑦 2
(a) Equation of contour curve 𝑪
A contour curve is defined by:
𝑓(𝑥, 𝑦) = constant
Level is:
𝑓(1,1) = 1 − 12 − 12 = 1 − 1 − 1 = −1
So:
1 − 𝑥 2 − 𝑦 2 = −1
Rearrange:
𝑥2 + 𝑦2 = 2
Answer:
𝐶: 𝑥 2 + 𝑦 2 = 2
(b) Vector perpendicular to 𝑪
Perpendicular vector = gradient:
∇𝑓 = (−2𝑥, −2𝑦)
At (1,1):
∇𝑓(1,1) = (−2, −2)
Answer:
(−2, −2)
