A microscopic
bug is traveling along the surface whose equation is x2 +
4y2 + 16z2 = 169.
(Assume
distances are measured in centimeters, so the ellipsoid is 26 cm wide,
13 cm deep, and 13/2 cm high, and times are measured in seconds.)
At time
t=0,
the bug is at the point (3,-2, 3), and traveling so that its x-coordinate
is increasing at the rate of 1/16 cm/sec and its y-coordinate is increasing
at the rate of 1/8 cm/sec.
How fast
is its z-coordinate changing at that instant?
There
are lots of reasonable strategies for attacking this question. Write some
down now.
1. Solve
for z in terms of x and y.
(The calculator
will give you more than one choice; make sure you use the right one.)
Then you're
stuck.
You can
figure out what to do next by reading Stewart's chapter 11.
By the end
of this project you'll have a quick-and-dirty answer.
For now,
try something different.
2. Use linear
approximations for the bug's x- and y-coordinates.
The bug's
x-coordinate is increasing 1/16 cm/sec and its x-coordinate is 3 at t=0,
so x = 3 + t/16, approximately.
The bug's
y-coordinate is increasing 1/8 cm/sec and its y-coordinate is -2 at t=0,
so y = -2 +t/8 , approximately.
Substitute
those expressions for x and y, then you can solve
for z and compute dz/dt
directly.
(You'll
still get two choices when you solve for z.)
3. You can
substitute x = 3 + t/16 and y = -2 +t/8 and use implicit
differentiation to find dz/dt.
4. (My favorite.)
Use linear approximations for the bug's x-, y-, and z-coordinates:
Suppose
dz/dt
is some unknown value m (cm/sec) at t=0, so that z = 3
+ m*t , approximately.
Substitute3
+ t/16 for x, -2 + t/8 for y, and 3 + m*t
for z in the equation of the ellipsoid and solve
the resulting equation for m.
(It's quickest
to differentiate the equation with respect
to t first.)
When you
solve for m, does the answer have t's in it?
What value
of t is the important one?
Question:
Do
you get the same answers if the bug is traveling along a line tangent to
the ellipsoid instead of along the ellipsoid itself? Does that make sense?
5. You will
be able to justify this technique later in your calculus career.
For now,
just do the computation to see if you get the same answer.
Use implicit
differentiation to find the rate of change of z with respect
to x at (3,-2,3), and multiply by 1/16.
(Ignore
y temporarily.)
This gives
the rate of change of z due to the change in x.
Next, use
implicit
differentiation to find the rate of change of z with respect
to y at the same point, ignoring
x temporarily, and multiply
by 1/8.
This gives
the rate of change of z due to the change in y.
Add these
two numbers.
Exercises: 1. Apply the last four of these strategies to the bug problem.
2. Apply
(more than one of) these strategies to related rates problems in the textbook.
These techniques
will work with problems in which more than two quantities are changing.