8—Multivariable Calculus 206Problems8.1 Letr=
√
x^2 +y^2 ,x=Asinωt,y=Bcosωt. Use the chain rule to compute the derivative
with respect totofekr. Notice the various checks you can do on the result, verifying (or disproving)
your result.
8.2 Sketch these functions* in plane polar coordinates:
(a)r=acosφ (b)r=asecφ (c)r=aφ (d)r=a/φ (e)r^2 =a^2 sin 2φ
8.3 The two coordinatesxandyare related byf(x,y) = 0. What is the derivative ofywith respect
toxunder these conditions? [What isdfalong this curve? And have you drawn a sketch?] Make
up a test function (with enough structure to be a test but still simple enough to verify your answer
independently) and see if your answer is correct. Ans:−(∂f/∂x)
/
(∂f/∂y)
8.4 Ifx=u+vandy=u−v, show that
(
∂y
∂x
)
u=−
(
∂y
∂x
)
vDo this by application of the chain rule, Eq. (8.6). Then as a check do the calculation by explicit
elimination of the respective variablesvandu.
8.5 Ifx=rcosφandy=rsinφ, compute
(
∂x
∂r
)
φand(
∂x
∂r
)
y8.6 What is the differential off(x,y,z) = ln(xyz).
8.7 Iff(x,y) =x^3 +y^3 and you switch to plane polar coordinates, use the chain rule to evaluate
(
∂f
∂r
)
φ,
(
∂f
∂φ
)
r,
(
∂^2 f
∂r^2
)
φ,
(
∂^2 f
∂φ^2
)
r,
(
∂^2 f
∂r∂φ
)
Check one or more of these by substitutingrandφexplicitly and doing the derivatives.
8.8 When current I flows through a resistance Rthe heat produced is I^2 R. Two terminals are
connected in parallel by two resistors having resistanceR 1 andR 2. Given that the total current is
divided asI=I 1 +I 2 , show that the condition that the total heat generated is a minimum leads to
the relationI 1 R 1 =I 2 R 2. You don’t need Lagrange multipliers to solve this problem, but try them
anyway.