**Unformatted text preview: **Math 210 DHC
Kouba
Discussion Sheet 2 1 (927‘ 82T
satisﬁes the equation — + —— 2 T3. «1:2 + y2 82:2 By? 2.) Find 21, zwzzmzyy, and 29:3, for z = ln(:L‘y2 + 3). 1.) Show that T = 3.) Find a function 2 = f (Lt, y) with the following partial derivatives or state
that this is impossible : zz 2 6’723’ cosa: + 2mye$2y sina: + 2my3 + 1 2 . 2
zy = 1228”” y sma: + 3cc2y2 + Zyey . 8 8
4.) Assume that u = f(m,y), x = rcosd, and y = rsin6. Compute 8—1:, 5;,
(92a
and W. 5.) Find an equation of the plane tangent to the surface z = y2 — 51:2 at the
point (2,1,—3) . 6.) Find the point on the plane 3:3 + 2y + 2 = 12 which is nearest the origin. 7.) a.) Show that (0,0) is a critical point for z = x4 — 2x2y2 + y“. Show that
(0,0) determines a minimum value. b.) Show that (0, 0) is a critical point for 2 = 21:4 + 4131/ + 11“. Show that
(0, 0) determines a saddle point. c.) Show that (0,0) is a critical point for z = (y — m2)(y — 2532). Show
that (0, 0) determines a saddle point. 9.) A house in the shape of a rectangular box is to hold 10,000 cubic feet. The
four walls admit heat at 5 units per minute per square foot. The roof admits
heat at 3 units per minute per square foot. The ﬂoor admits heat at 2 units
per minute per square foot. What should the dimensions (length, width, height) of the house be in order to minimize the rate (units per minute) at
which heat enters ? ...

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- Spring '09
- Kouba
- Calculus