Math 53: Worksheet 5

September 26

1. Sketch the following surfaces.

(a) y2 + 4 = x2 + 4z2.

This equation represents a hyperboloid of one sheet wrapped around the y-axis.

-2

-1

0

1

2

-2.6904

Z a

xis

-1.61424

5.38516

-0.53808

0.53808

3.23459

1.61424

2.6904

1.084025

3X axis

-1.066561

-1 -3.21713Y axis

-3

-5.3677-5

(b) y2 + 2z2 = 6.

This equation represents a cylinder wrapped around the x-axis. Note that thecylinder will have a radius of

√6 in the y direction and

√3 in the z direction.

-1.5

-1

-0.5

0

0.5

1

1.5

-1.73065

Z a

xis

-1.03839

-0.346129

-5

0.346129

-2.44155-3

1.03839

-1.46334

1.73065

-1-0.485132

X axis

10.493075Y axis

31.47128

52.44949

1

(c) x2 − y2 − z2 = 4.

We have x2 = 4 + y2 + z2. This represents a hyperboloid of two sheets about thex-axis.

-4

-2

0

2

4

-5.63851-5.65227

-3.39136

-3.37944 -6

-1.13045

1.13045

-3.6-1.12036

Z a

xis

3.39136

-1.21.13871

5.65227

Y axis

1.2 X axis

3.397783.6

5.65685 6

2. Find a vector function that represents the curve of intersection of the cylinder x2+y2 =9 and the plane x+ 2y + z = 3.

Note that the cylinder can be parametrized as x = 3 ∗ cos(t), y = sin(t), where0 ≤ t < 2π, with z ∈ R. Plugging these in the equation of the plane gives

z = 3− x− 2y = 3− 3 cos(t)− 6 sin(t).

The curve of intersection is therefore given by

r(t) =< 3 cos(t), 3 sin(t), 3− 3 cos(t)− 6 sin(t) > , 0 ≤ t < 2π.

3. Find the point of intersection of the tangent lines to the curve r(t) =< sin(πt), 2 sin(πt), cos(πt) >at the points where t = 0 and t = 0.5.

We haver′(t) =< π cos(πt), 2π cos(πt),−π sin(πt) > .

At t = 0 and t = 0.5, we get v := r′(0) =< π, 2π, 0 > and w := r′(0.5) =< 0, 0,−π >.Also let a := r(0) =< 0, 0, 1 > and b := r(0.5) =< 1, 2, 0 >. The two tangent lineshave equations

r1(s) = a + sv =< πs, 2πs, 1 >

r1(u) = b + uw =< 1, 2,−πu >where s, u ∈ R. At the point of intersection, we require

πs = 1, 2πs = 2, 1 = −πu

so that s = 1π

and u = − 1π. The point of intersection is therefore

r1(1/π) = r2(−1/π) = < 1, 2, 1 > .

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4. Show that the curve with parametric equations x = sin(t), y = cos(t), z = sin2(t) isthe curve of intersection of the surfaces z = x2 and x2 + y2 = 1. Use this fact to helpsketch the curve.

Note that the parametric equations of the curve obey both z(t) = x(t)2 and x(t)2 +y(t)2 = 1 for all values of t.

0

0.1

0.2

1

0.3

0.4

0.5

0.6

z-a

xis

0.7

0.8

0.5

0.9

1

1

y-axis

0 0.5

x-axis

0-0.5-0.5

-1 -1

5. Draw contour maps of the following functions showing several level curves.

(a) f(x, y) = y sec(x).

Observe that f(x, y) = k ⇒ y sec(x) = k ⇒ y = k cos(x). The level curves off are therefore just cosine waves of differing amplitudes. Note that the pointswhere the level curves meet are not actually part of the domain of f .

0 1 2 3 4 5 6 7

x

-3

-2

-1

0

1

2

3

y

(b) f(x, y) = xx2+y2

.

3

Let k 6= 0 first. Observe that f(x, y) = k ⇒ 1kx = x2 + y2 ⇒ (x− 1

2k)2 + y2 = 1

4k2.

The level curves of f are therefore just circles centered at different points on thex-axis and passing through the origin. Note again that the origin is not actuallypart of the domain of f . For k = 0, we simply get f(x, y) = 0 ⇒ x = 0, i.e, they-axis.

-1 -0.5 0 0.5 1

x

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

y

(c) f(x, y) = sin(xy).

Let −1 ≤ k ≤ 1. Observe that f(x, y) = k ⇒ sin(xy) = k ⇒ xy = arcsin(k) ⇒y = arcsin(k)

x. The level curves of f therefore are the inverse proportionality curves

between x and y.

-3 -2 -1 0 1 2 3

x

-3

-2

-1

0

1

2

3

y

6. Find the limit, if it exists, or show that the limit does not exist.

(a) lim(x,y)→(0,0)5x2yx2+y2

.

We show that the limit is zero. Switching to polar coordinates gives

lim(x,y)→(0,0)

5x2y

x2 + y2= lim

r→0

5r3 cos2(θ) sin(θ)

r2= lim

r→05r cos2(θ) sin(θ).

4

Since 5 cos2(θ) sin(θ) is bounded, we conclude that the limit is zero.

(b) lim(x,y)→(0,0)sin(xy2)x2+y4

.

Along the y-axis, we have x = 0 so we get limy→0sin(0)y4

= 0. Along the curve

x = y2, however, we get limy→0sin(y4)2y4

= 12. Since we get different values along

different paths, we conclude that the required limit does not exist.

(c) lim(x,y)→(0,0)2x

x2+x+y2.

Along the y-axis, we have x = 0 so we get limy→00y2

= 0. Along the x-axis, we

have y = 0 so we get limx→02x

x2+x= 2. Since we get different values along different

paths, we conclude that the required limit does not exist.

(d) lim(x,y)→(0,0)x2+y2√

x2+y2+1−1.

Note that rationalization yields

lim(x,y)→(0,0)

x2 + y2√x2 + y2 + 1− 1

×√x2 + y2 + 1 + 1√x2 + y2 + 1 + 1

= lim(x,y)→(0,0)

(x2 + y2)(√x2 + y2 + 1 + 1)

x2 + y2 + 1− 1

whence we get lim(x,y)→(0,0)

√x2 + y2 + 1 + 1. Since this function is continuous at

(0, 0), the limit is√

0 + 0 + 1 + 1 = 2.

(e) lim(x,y)→(0,0)e−x2−y2−1x2+y2

.

Switching to polar coordinates gives

lim(x,y)→(0,0)

e−x2−y2 − 1

x2 + y2= lim

r→0

e−r2 − 1

r2= lim

r→0

−2re−r2

2r= lim

r→0−e−r2 = −1

where we used L’Hopital’s rule.

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