Solutions to Homework 5

13.3

#3 z_{x} = x and z_{y} = 4y, so z(2, 1) = 4, z_{x}(2, 1) = 2 and z_{y}(2, 1) = 4. Therefore the equation of the tangent plane at (x, y) = (2, 1) is z = z(2, 1) + z_{x}(2, 1)(x − 2) + z_{y}(2, 1)(y − 1) = 4 + 2(x − 2) + 4(y − 1) or

z = −4 + 2x + 4y.

#4 [4]

a) The answer is wrong because it involves a quadratic term (x^{2}) and tangent planes are planes (linear functions) and can therefore only involve linear (mx, ny) and constant terms (c).

b) He did not evaluate the partial derivatives!

c) f_{x}(2, 3) = 3(2)^{2} = 12 and f_{y}(2, 3) = −2(3) = −6, so that the answer is

z = 12(x − 2) − 6(y − 3) − 1 = 12x − 6y − 7.

#15

dg = (2u + v)du + (u)dv

#23

dA = ydx + xdy and so |dA| ≤ 6 · 0.01 + 5 · 0.01 = 0.11 square feet.

#25

dl = \frac{∂l}{∂s}ds + \frac{∂l}{∂k} dk = (1 − (k^{2}/s^{2}))ds + (2k/s)dk.

Here a 1% error means |Δs| = 0.01s and |Δk| = 0.01k. So

Δl ≈ (1 − (k^{2}/s^{2}))0.01s + (2k/s)0.01k = 0.01[s − k^{2}/s + 2k^{2}/s] =
0.01[s + k^{2}/s] = 0.01l.

So the maximum error in l is 1%. Next,

dg = \frac{∂g}{∂l} dl + \frac{∂g}{∂T} dT = \frac{4π^{2}}{T^{2}} dl − \frac{8π^{2}l}{T^{3}} dT.

Notice that the maximum error occurs when Δl and ΔT are opposite signs, and the maximum error on g is

Δg ≈ \frac{4π^{2}}{T^{2}} |Δl| + \frac{8π^{2}l}{T^{3}} |ΔT | =

\frac{4π^{2}}{T^{2}} 0.01l + \frac{8π^{2}l}{T^{3}} 0.005T = 0.02\frac{4π^{2}l}{T^{2}} = 0.002g.

So the maximum error on the computed value of g is 2%.

13.4

#1 Here the displacement vector is < 3 − 1, 5 − 4 >=< 2, 1 >, so normalizing this vector we get u⃗ =< 2/\sqrt{5}, 1/\sqrt{5} > is a unit vector in the direction in which we are traveling.

a)

fu⃗(1, 2) ≈ \frac{f(1 + h2/\sqrt{5}, 4 + h1/\sqrt{5}) − f(1, 4)}{h}

for small values of h. For example h = 0.1 gives the value is ≈ 1.0, which is fairly good, but usually we should take smaller values of h to get more accuracy.

b)

fu⃗(3, 5) ≈ \frac{f(3, 5) − f(3 − h2/\sqrt{5}, 5 − h1/\sqrt{5})}{h}

for small values of h and again h = 0.1 yields a reasonable estimate of 0.98.

#9 Both are negative since the function is decreasing in both of those directions at (2,5). More specifically, the y-direction has no influence on whether the function values increase or decrease, but for the given x = 2 we get that moving in positive x-direction, as we do in both cases, decreases the z-value.

#12 First normalize v⃗. ∥v⃗∥ = 5, so that u⃗ =< 4/5, −3/5 > is a unit vector in the direction of v⃗. f_{x} = 2xy and f_{y} = x^{2}, so that

fu⃗(2, 6) = f_{x}(2, 6)(4/5) + f_{y}(2, 6)(−3/5) = (24)(0.8) + (4)(−0.6) = 16.8

#13 [3] Similar to #12. u⃗ =< 2/3, 1/3, −2/3 > and f_{x} = 2x + 3y, f_{y} = 3x and f_{z} = 2. So fu⃗(2, 0, 1) = 4(2/3) + 6(1/3) + 2(−2/3) = 10/3

#14

Same procedure as in #13:

a) u⃗_{1} =< 1/\sqrt{2}, 0, −1/\sqrt{2} >, so fu⃗_{1}(−1, 0, 4) = 0(1/\sqrt{2}) + 8(0) + 0(−1/\sqrt{2}) = 0.

b) u⃗_{2} =< −1/\sqrt{19}, 3/\sqrt{19}, 3/\sqrt{19} >, so fu⃗_{2}(−1, 0, 4) = 0(−1/\sqrt{19}) + 8(3/\sqrt{19}) + 0(3/\sqrt{19}) = 24/\sqrt{19}.

13.5

#1 True, take the direction perpendicular to gradf at that point; this is the same as going along the level curve (neither going up or down). If gradf=0, then the function is flat at that point and you can go in any direction.

#18

∇z = z_{x}i⃗ + z_{y}j⃗ = \frac{ye^{y}}{(x + y)^{2}} i⃗ + \frac{e^{y}(x^{2} + xy − x)}{(x + y)^{2}}j⃗.

#31 [5] v⃗_{1} =< −1, 2 > and v⃗_{2} =< 3, 4 > are the displacement vectors from (2,1) to (1,3) and (5,5) respectively. Normalize them to get u⃗_{1} =< −1/\sqrt{5}, 2/\sqrt{5} > and u⃗_{2} =< 3/5, 4/5 >. So we know that fu⃗_{1} = −2/\sqrt{5} and fu⃗_{2} = 1. Also let f_{x}(2, 1) = a and f_{y}(2, 1) = b, so that the gradient of f is ∇f(2, 1) = ai⃗+bj⃗. But also we know for the directional derivatives, that fu⃗ = ∇f · u⃗, so that we get that:

−2/\sqrt{5} = a(−1/\sqrt{5}) + b(2/\sqrt{5}) from u⃗_{1},
and 1 = a(3/5) + b(4/5) from u⃗_{2}.

Solving this system of linear equations we get that a = 1.8 and b = −0.1.

#34

a) ∇F = (2x + 2xz^{2})i⃗ + (4y^{3})j⃗ + (2x^{2}z)k⃗, and evaluating at (-1,1,1) we get that we should move in the direction of steepest ascent of concentration (namely gradf), which is −4i⃗ + 4j⃗ + 2k⃗.

b) ∥∇F ∥ = \sqrt{16 + 16 + 4} = 6, so that is the rate at which the concentration is changing in the direction of a) with respect to distance. You also know that the distance is changing at a rate of 4 units/sec. So (using the chain rule) we get that:

change(conc/time) = [change(conc/dist)][change(dist/time)]

= [6(unitsofconc/unitsofdistance)][4(unitsofdist/sec)]

= 24(unitsofconc/sec).

#38 In either case determine when gradf=gradz, since this is the only way how the tangent plane can be parallel to the given plane. First compute gradf = 2xi⃗+2yj⃗.

a) gradz = 0, so that the gradients are equal exactly when x = y = 0. So the planes are parallel only for the point (0,0,1) on the surface.

b gradz = 6i⃗− 10j⃗, so that the gradients are equal exactly when x = 3 and y = −5. So the planes are parallel only for the point (3,-5,35) on the surface.

13.6

#5

\frac{dz}{dt} = \frac{∂z}{∂x} \frac{dx}{dt} + \frac{∂z}{∂y} \frac{dy}{dt} = (e^{y})(2) + (xe^{y})(−2t) = 2e^{1−t^{2}}(1 − 2t).

#10 [4]

\frac{∂z}{∂u} = \frac{∂z}{∂x} \frac{∂x}{∂u} + \frac{∂z}{∂y} \frac{∂y}{∂u} = (e^{y})(1/u) + (xe^{y})(0) = e^{v}/u.

\frac{∂z}{∂v} = \frac{∂z}{∂x} \frac{∂x}{∂v} + \frac{∂z}{∂y} \frac{∂y}{∂v} = (e^{y})(0) + (xe^{y})(1) = e^{v} ln u.

#15

\frac{∂z}{∂u} = \frac{∂z}{∂x} \frac{∂x}{∂u} + \frac{∂z}{∂y} \frac{∂y}{∂u} = (e^{y})(2u) + (xe^{y} + e^{y} + ye^{y})(2u) = 2ue^{u^{2}−v^{2}}(2u^{2} + 2).

\frac{∂z}{∂v} = \frac{∂z}{∂x} \frac{∂x}{∂v} + \frac{∂z}{∂y} \frac{∂y}{∂v} = (e^{y})(2v) + (xe^{y} + e^{y} + ye^{y})(−2v) = −4vu^{2}e^{u^{2}−v^{2}}.

#21 [4] The tree diagram has w at the top and branches out from there into x, y and z. From each of those it branches out to both u and v, so that the tree will have 3 u’s and 3 v’s on the bottom.

\frac{∂w}{∂u} = \frac{∂w}{∂x} \frac{∂x}{∂u} + \frac{∂w}{∂y} \frac{∂y}{∂u} + \frac{∂w}{∂z} \frac{∂z}{∂u}.

\frac{∂w}{∂v} = \frac{∂w}{∂x} \frac{∂x}{∂v} + \frac{∂w}{∂y} \frac{∂y}{∂v} + \frac{∂w}{∂z} \frac{∂z}{∂v} .