11Ma1cPracHW1Sol

7/30/2019 11Ma1cPracHW1Sol

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SOLUTIONS TO HOMEWORK 1

MA1C PRACTICAL 2011-2012

(1) The level curves are determined by the equations x y + 2 = c; the level curveof height c is the line y = x + 2 c. Here is a picture of the level curves withc {3, 2, . . . , 2, 3}:

The graph is a plane in R3, and it looks like this:

(2) The level curves are determined by the expression x2 + 4y2 = c, for each constant

c R. If c < 0, there are no (x, y) R2 satisfying the expression, and if c = 0,the only such point is (0, 0). Ifc > 0, such points form an ellipse centered at the

1


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origin with semimajor axis

c and semiminor axis

c/2. The graphics below depict,

respectively, some level curves of f and part of the graph of f.

(3) Note that x2 + xy = c y = c/x x when x = 0. Thus the level curves will behyperbolas with asymptotes at the lines x = 0 and y = x. Here are the level curvesfor f(x, y) = x2 + xy at c = 3, 2, 1, 0, 1, 2, 3:


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(4) The graph is a circular cone about the y-axis:

(5) (a)

lim(x,y)(0,0)

(x + y)2 (x y)2xy

Note that (x + y)2 = x2 + 2xy + y2 and (x y)2 = x2 2xy + y2, so we have

lim(x,y)(0,0)

(x + y)2

(x y)2

xy= lim

(x,y)(0,0)4xyxy

= lim(x,y)(0,0)

4

= 4.


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(b)

lim(x,y)(0,0)

sin xy

y

Note that for x = 0 we have xx

= 1, so we can write

lim(x,y)(0,0)

sin xyy = lim(x,y)(0,0) x sin xyxy

= lim(x,y)(0,0)

xsin xy

xy

and thus if lim(x,y)(0,0)sinxyxy

exists then we have

lim(x,y)(0,0)

sin xy

y= ( lim

(x,y)(0,0)x)( lim

(x,y)(0,0)

sin xy

xy)

= 0cdot( lim(x,y)(0,0)

sin xy

xy)

= 0.

But let t = xy. Then we have

( lim(x,y)(0,0)

sin xy

xy) = (lim

t0

sin t

t)

and we know that limt0sin tx

= 1 (e.g. by LHopitals Rule), so the limit exists

and we have

lim(x,y)(0,0)

sin xy

y= 0.

(Alternatively, we can say: since limt0sin tt

= 1, whenever we have an > 0 we

can pick > 0 such that |t| < implies | sin tt

1| < . Then set =

and when

(x, y) < then |x|, |y| < so |xy| < 2 = , and we have | sin(xy)xy

1| < asdesired.)

(c)

lim(x,y)(0,0)

x3 y3x2 + y2

First note that x2

x2 + y2 and y2

x2 + y2, and thus 1

x2+y2

1x2

, 1y2

. Then we

have

x3 y3x2 + y2

|x

3|x2 + y2

+|y3|

x2 + y2

|x|3

x2+

|y|3y2

= |x| + |y|.


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Thus we get 0 x3y3

x2+y2

|x|+ |y|. But lim(x,y)(0,0) 0 = lim(x,y)(0,0) |x|+ |y| = 0

and so by the squeeze theorem, we have

lim(x,y)(0,0)

x3 y3x2 + y2

= 0.

(d)

limx0

sin2x 2xx3

Solution. Since sin 2x 2x and x3 approach 0 as x 0, we can hope to uselHopitals rule. The derivatives 2 cos 2x 2 and 3x2 also each approach 0, asdo the second derivatives 4sin2x and 6x. The ratio of the third derivatives,(8cos2x)/6, equals -4/3, so our hope is fulfilled. By lHopitals rule, the limitof the original function as x 0 is -4/3.

(e)

lim(x,y)(0,0)

sin(2x) 2x + yx3 + y

Solution. The limit does not exist, as we can see by approaching (0, 0) along two

well-chosen lines. If we take (x, y) to (0, 0) along the line x = 0, the values of the

function approach

limy0

y

y= lim

y01 = 1.

If we approach (0, 0) along the line y = 0, the values of the function approach

limx0

sin2x 2xx3

,

which by (a) is equal to 4/3. Since these two limits are unequal, the two-variablelimit does not exist.

(f)

lim(x,y,z)(0,0,0)

2x2y cos z

x2 + y2

Solution.Since 0 | cos z| 1, we have the bound 0 |

2x2y cos z

x2+y2 | |2x2y

x2+y2 |.And | 2x2y

x2+y2| |2x2y

x2| = 2|y|, which approaches 0 as (x,y,z) (0, 0, 0). So by the

squeeze theorem,

lim(x,y,z)(0,0,0)

2x2y cos z

x2 + y2= 0.


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(6) Note that f(x, y) is not defined on the line y = x. We want to know if its possible toextend f(x, y) to a continuous function on the domain R2 \ {(x, y) : x = y} (0, 0).Since x + y 0 as x 0 and y 0, it follows from limx+y0[sin(x + y)]/(x + y) = 1that limx0,y0[sin(x + y)]/(x + y) = 1. So the function can be made continuous on

its new domain, R2 \ {(x, y) : x = y} (0, 0), by defining f(0, 0) = 1.

(7) Suppose y = cx for some constant c. Then for x = 0,xy

x2 + y2=

cx2

x2 + c2x2=

c

1 + c2.

Taking c = 0 and then c = 1 gives c1+c2 = 0 and 1/2 respectively, so the limit of the

function does not exist at (0, 0). So the function cannot be made continuous at (0, 0).

(8) We know that the functions x,y,ex, sin x : R R are continuous. It is then immedi-ate from the definition of continuity that they are continuous as functions R2 R.Then by parts (ii) and (iii) of Theorem 4 in section 2.2 (continuity is preserved by

addition and multiplication) , the function yex + sin x + (xy)4 is continuous.

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11Ma1cPracHW1Sol