The University of SydneySchool of Mathematics and Statistics

Solutions to Tutorial for Week 8

MATH1011: Applications of Calculus Semester 1, 2014

Web Page: http://www.maths.usyd.edu.au/u/UG/JM/MATH1011/Lecturer: Clinton Boys, Ross Ogilvie and George Papadopoulos

In this tutorial, you should also complete the questions fromTutorial for Week 7 which you missed due to last weeks quiz.

Question to complete BEFORE the tutorial

1. Find both first order partial derivatives for the following functions:

(a) z = 5x2 6x2y + y3 + 2.Solution: zx = 10x 12xy, zy = 6x2 + 3y2.

(b) z = (x2 + 2xy + y2) cos y + x4 sin y.

Solution:

zx = (2x + 2y) cos y + 4x3 sin y.

zy = (2x + 2y) cos y + (x2 + 2xy + y2)( sin y) + x4 cos y

= (x4 + 2x + 2y) cos y (x2 + 2xy + y2) sin y.(c) z = ln(25x3y4) + ln(y4).

Solution: First simplify the expression using the log laws:

ln(25x3y4) + ln(y4) = ln(25x3y4y4) = ln(25x3).

Now differentiate using the chain rule, noticing that the dependence on y has gone:

zx =1

25x3 (75x2) = 3

x.

zy = 0.

(d) f(x, y) =e3+2y

3x + 5.

Solution: fx =3e3+2y(3x + 5)2

, fy =2e3+2y

3x + 5.

Tutorial questions

2. (a) Let f(x, y) = x2 + 3xy y2 + x 5y + 25. Find all the points of the surfacez = f(x, y) at which the tangent planes to the surface are horizontal.

Solution: Calculate the first partial derivatives: fx = 2x+3y+1, fy = 3x2y5.The tangent planes are horizontal at the points where both derivatives vanish:

2x + 3y + 1 = 0

3x 2y 5 = 0Solving the equations simultaneously we find x = 1 and y = 1. Thus, (1,1, 28)is the only point at which the tangent plane is horizontal.

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(b) The curves C and T are formed by cutting the paraboloid z = x2 + y2 7 withthe planes x = 2 and z = 9 respectively. Write down the equations for C and T .

Solution: The equation of the curve C is z = y2 3. The equation of T isx2 + y2 = 16.

3. For the functions below and the regions on which they are defined, state whether thefunction has a global maximum and minimum, and find them if so.

(a) The paraboloid z = 4 + x2 + y2, defined on the whole (x, y)-plane.

Solution: The infinite region on which the function is defined has no boundary.The partial derivatives are fx = 2x and fy = 2y which are always defined. Hencea global maximum or minimum can only occur when fx = fy = 0, i.e. whenx = y = 0. This is the global minimum since f(x, y) = 4+(x2+y2) and x2+y2 0.Global minimum at (0, 0, 4), no global maximum.

(b) The paraboloid z = 4 + x2 + y2, defined on the region x2 + y2 25.Solution: Same as above, except we now also need to check the function valueson the boundary x2+y2 = 25. At these points, f(x, y) = 4+(x2+y2) = 4+25 = 29,which gives the global maximum.

Global minimum at (0, 0, 4), global maximum at (5, 5, 29).

(c) The plane z = 2x + 3y 2, defined on the whole (x, y)-plane.Solution: No boundary to the region on which the function is defined, and bothpartial derivatives (fx = 2, fy = 3) are always defined and never zero, so no globalmaximum or minimum.

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