Solutions to Homework 4

12.4

#8

k⃗ × r⃗ =< 0, 0, 1 > × < x, y, z >= −yi⃗ + xj⃗.

#15 [3] n⃗_{1} = 2i⃗−3j⃗+5k⃗ and n⃗_{2} = 4i⃗+j⃗−3k⃗. The line of intersection is perpendicular to both normal vectors, hence it is in the direction of n⃗_{1} × n⃗_{2} = 4i⃗ + 26j⃗+ 14k⃗ = n⃗.

#16 [3] n⃗ of problem 15 is the normal vector for the plane through (0,0,0) we are looking for. So its equation is 4(x − 0) + 26(y − 0) + 14(z − 0) = 0, and this simplifies to 4x + 26y + 14z = 0.

#19 We need a⃗ · v⃗ = 0 and b⃗ · v⃗ = 0. So this gets us two equations:

a_{1}x + a_{2}y + a_{3}z = 0(I)
b_{1}x + b_{2}y + b_{3}z = 0(II)

Multiply (I) by b_{1} and subtract a_{1} times (II) from it, to get

(b_{1}a_{2} − a_{1}b_{2})y + (b_{1}a_{3} − a_{1}b_{3})z = 0 or y = \frac{−(b_{1}a_{3} − a_{1}b_{3})z}{(b_{1}a_{2} − a_{1}b_{2})} for b_{1}a_{2} ≠ a_{1}b_{2}.

Next multiply (II) by a_{2} and subtract b_{2} times (I) from it, to get

(b_{1}a_{2} − a_{1}b_{2})x + (a_{2}b_{3} − b_{2}a_{3})z = 0 or x = \frac{−(a_{2}b_{3} − b_{2}a_{3})z}{(b_{1}a_{2} − a_{1}b_{2})} for b_{1}a_{2} ≠ a_{1}b_{2}.

Picking z = a_{1}b_{2} − b_{1}a_{2} yields x = a_{2}b_{3} − b_{2}a_{3} and y = b_{1}a_{3} − a_{1}b_{3}, which is the same result as we get for the algebraic method for computing the cross product.

13.1

#2 [3]

a) If you borrow $8,000 at an interest rate of 1% per month and pay it off in 24 months, then your monthly payments are $376.69.

b) The increase in your monthly payments for borrowing an extra dollar under the same terms as in a) is about 4.7 cents.

c) If you borrow the same amount of money for the same period of time as in a), but the interest rate increases by 1%, then your monthly payments will increase by roughly $44.83.

#9 f_{w}(10, 25) ≈ (f(15, 25) − f(5, 25))/10 = −1.9. This means that the apparent temperature drops by about 1.9^{o}F as the wind increases by 1mph. f_{T} (5, 20) ≈ (f(5, 25) − f(5, 15))/10 = .9. This means that the apparent temperature increases by about .9^{o}F as the outside temperature increases by 1^{o}F.

#15

a) \frac{∂P}{∂r} (8, 4000) ≈ (P (16, 4000) − P (8, 4000))/(16 − 8) = 2.5 in $ per %, and \frac{∂P}{∂L} (8, 4000) ≈ (P (8, 5000) − P (8, 4000))/(5000 − 4000) = 0.02 in $ per $ with the explanations being similar to the ones in #2c) and b).

b) \frac{∂P}{∂r} (8, 6000) ≈ 3.33 and \frac{∂P}{∂L} (8, 6000) ≈ 0.02.

c) \frac{∂P}{∂r} (13, 7000) ≈ 3.33 and \frac{∂P}{∂L} (13, 7000) ≈ 0.02, so roughly the same as in b). #21 [4]

a) f_{c}(c, s) = change in blood pressure as cardiac output is increased (while systemic vascular resistence (SVR) remains constant.

b) This should look like curves of the form s = p/kc, (here p,k constant, c independent variable, s dependent variable), so stretched y = 1/x’s, that is hyperbolas.

c) A and B should be on the same contour with A being to the left of B, since A has lower c and higher s than B.

d) Start on a high level curve with point D, drop to E on a lower level curve by going to the left in the diagram and then return to the original level curve by moving vertically up to F.

13.2

#1 a) f_{u}(1, 3) ≈ (f(1.001, 3) − f(1, 3))/.001 = 41.1,

b) f_{u} = (u^{2} +v^{2})^{3/2} +3u^{2}(u^{2} +v^{2})^{1/2} and thus f_{u}(1, 3) ≈ 41.11, so that our estimate

from a) was quite good.

#13

\frac{∂}{∂T} (2πrT^{−1}) = −2πrT^{−2}.

#21

\frac{∂}{∂M} (2πr^{3/2}(GM)^{−1/2}) = −πr^{3/2}(GM)^{−3/2}G.

#27 [2]

\frac{∂}{∂θ} (3t cos(5θ − 1) − tan(7θ^{2})) = −15t sin(5θ − 1) − 14tθ sec^{2}(7θ^{2}).

#28

\frac{∂}{∂x}(xe^{\sqrt{xy}}) = … = e^{\sqrt{xy}}(1 + .5\sqrt{xy}).

#31

\frac{∂m}{∂v} = −.5m_{0}(1 − v^{2}/c^{2})^{−3/2}(−2v/c^{2}).

#34

\frac{∂}{∂w} ([(w − 1)/(x^{2}yw − xy^{3}w^{7})]^{3.5}) = 3.5[(w − 1)/(x^{2}yw − xy^{3}w^{7})]^{2.5}[1 ∗ (x^{2}yw −
xy^{3}w^{7}) − (w − 1) ∗ (x^{2}y − 7xy^{3}w^{6})]/(x^{2}yw − xy^{3}w^{7})^{2}.

#36

\frac{∂H}{∂t} = \frac{1}{e^{t+p}+t}(e^{t+p} + 1) + (cos p^{2}t)p^{2}.

#38

\frac{∂f}{∂x} = ln(y cos x) + x\frac{1}{y cos x}(−y sin x), so that \frac{∂f}{∂x}|_{(π/3,1)} ≈ −2.51.

#39

K \frac{∂Q}{∂K} = bαK^{α}L^{1−α}, and L\frac{∂Q}{∂L} = b(1 − α)K^{α}L^{1−α}, so that altogether, K \frac{∂Q}{∂K} +
L\frac{∂Q}{∂L} = b(α + 1 − α)K^{α}L^{1−α} = bK^{α}L^{1−α} = Q.