Math 120 Answers for Homework 24 1. Equation Picture Possible Reasons (a) z = cos x 2 + y 2 -8 -1 -8 -4 -4 -0.5 0 0 x y 4 4 0 8 8 0.5 1 – Graph is rotationally symmetric – Height at all maxima is 1 – Cross section is cos graph. (b) z = sin(y ) -4 -2 -1 -4 0 x -0.5 -2 2 0 0 y 2 0.5 4 4 1 – Graph and function do not depend on x. – Slice when x constant is sin. (c) z = cos 2 (x) sin 2 (y ) -3 -2 0 -3 0.2 -1 -2 0.4 -1 0 0.6 x 0.8 0 1 y 1 1 2 2 3 3 – Graph is always positive. – Graph is periodic on each x-slice and y-slice. (d) z = sin(x 2 +y 2 ) x 2 +y 2 -3 -2 -1 0 y 3 -0.2 1 2 0 1 0 2 0.2 -1 x 0.4 -2 3 -3 0.6 0.8 1 – Graph is rotationally symmetric – Maximum height decreases with distance from origin. 1
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Math 120 Answers for Homework 24 - Queen's Universitymikeroth/oldteaching/Math120/… · · 2009-08-31Math 120 Answers for Homework 24 1. Equation Picture Possible Reasons (a) z
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Math 120 Answers for Homework 24
1.
Equation Picture Possible Reasons
(a) z = cos(
√
x2 + y2)
-8-1-8
-4-4
-0.5
00 x
y44
0
88
0.5
1
– Graph is rotationallysymmetric
– Height at all maxima is 1
– Cross section is cos graph.
(b) z = sin(y)
-4
-2-1
-40 x
-0.5
-2 2
0
0y 2
0.5
44
1
– Graph and function do notdepend on x.
– Slice when x constant is sin.
(c) z = cos2(x) sin2(y)
-3-20
-3
0.2
-1-2
0.4
-1 0
0.6
x
0.8
0 1y
1
1 2233
– Graph is always positive.
– Graph is periodic on eachx-slice and y-slice.
(d) z = sin(x2+y2)x2+y2
-3
-2
-1
0 y
3-0.2
12
0
10
2
0.2
-1x
0.4
-2 3-3
0.6
0.8
1
– Graph is rotationallysymmetric
– Maximum height decreaseswith distance from origin.
1
(e) z = cos(xy)
-3
-2
-1
0-1-3 x
-2 1-1
-0.5
0 21
0
y2 3
3
0.5
1
– Graph has height 1 whenxy = 0.
– For fixed x or y (i.e., x-slices ory-slices) graph looks like cos.
(f) z = |xy|
0 -2-2
1
-1-1
2
0
3
0 xy
4
11
22
– Slice when y constant looks likea multiple of |x|, and slice whenx constant looks like a multipleof |y|.
2. Starting with f(x, y) = xesin(y) + y2 we compute that
(a) ∂f
∂x= esin(y).
(b) ∂f
∂y= (cos(y)) · xesin(y) + 2y.
(c) ∂3f
∂x2y= 0
(the fastest way to see this is to differentiate ∂f
∂xwith respect to x to get ∂2f
∂x2 = 0,
and then differentiate once more with respect to y to get ∂3f
∂x3 = 0).
3. The tangent plane to a graph z = f(x, y) over the point (x0, y0) is the plane givenby the equation z = ax + by + c with a = fx(x0, y0), b = fy(x0, y0) and c chosen so thatthe tangent plane passes through the point (x0, y0, f(x0, y0)). This is the same as sayingthat c = f(x0, y0) − x0fx(x0, y0) − y0fy(x0, y0).
(c) h(x, y) = e(x2+y2)2 at the point (x0, y0) = (3, 4).
function value at (3, 4) constant
hx 4x(x2 + y2)e(x2+y2)2 300e625 a = 300e625
hy 4y(x2 + y2)e(x2+y2)2 300e625 b = 300e625
h e(x2+y2)2 e625 c = e625 − 3 · (300e625) − 4 · (300e625) = −2099e625
The equation of the tangent plane is therefore
z = 300e625x + 300e625y − 2099e625.
4.
(a) f(x, y) = x3 − y3 + 3xy.
f = x3 − y3 + 3xy
fx = 3x2 + 3y
fy = −3y2 + 3x
fxx = 6x
fyy = −6y
fxy = 3
∂∂x
∂∂y
∂∂x
∂∂y
∂∂x
∂∂y
3
At the point (x0, y0) = (1,−1) the value of the partial derivatives (and D) are :
Point fxx(x0, y0) fyy(x0, y0) fxy(x0, y0) D Classification(0, 0) 6 6 3 6 · 6 − 32 = 27 Min
Since D > 0 the point is either a local max or a local min. Since the secondderivative fxx is negative we see that (0, 0) is a local min by the rule for the onevariable test.
(b) g(x, y) = xy
g = xy
gx = y · xy−1
gy = ln(x) · xy
gxx = y · (y − 1) · xy−2
gyy = (ln(x))2 · xy
gxy = (1 + y ln(x)) · xy−1
∂∂x
∂∂y
∂∂x
∂∂y
∂∂x
∂∂y
At the point (x0, y0) = (1, 0) the values of the partial derivatives and D are: