Test
2 Notes
Functions of Several
Variables –
Section 11.1, P.756
à parabola
Partial Derivatives – Section 11.3, P.776
Definition: ![]()


some value (plugin x,
a, b, and c)
|
Chart |
y=15 |
y=20 |
y=25 |
y=30 |
|
x=10 |
4 |
5 |
5 |
5 |
|
x=20 |
7 |
8 |
9 |
9 |
|
x=30 |
13 |
16 |
18 |
21 |
|
x=40 |
21 |
25 |
31 |
31 |
|
x=50 |
29 |
36 |
40 |
45 |
1) Hold y constant at y=25; let b = 30 so: a = 40, c = 20 … find dz/dx:
![]()
2) Hold x constant at x=30; let b = 25 so: a = 30, c = 20 … find dz/dy:
![]()
Tangent Planes and
Linearization
– Section 11.4, P.788
![]()

![]()
Find
Linearization at base point (let x0 and y0 equal the
middle value b from above … so x0
= 30; y0 = 25)
![]()
Plugin
the x and y to find the approximate value at x, y:
![]()
Chain Rule – Section 11.5, P.796
Find
the pieces you need to get dz/dt à dz/dx, dx/dt, dz/dy, and
dy/dt:
![]()
![]()
![]()

o
3) Related Rates: you work
with some object’s volume, the rate of change of the volume and the variables
it depends on, and the initial values of those variables ... one of these
things will not be given and you’ll have to solve for it (typically rate of
volume change)
![]()
We’ve been
asked for dV/dt (rate volume is changing over time) so everything else must be
given:
dV/dr = rate of change of V with respect to r = fr(r,
h) = (partial deriv
of V w/resp. to r) = 2/3π r
h
dV/dh = rate of change of V with respect to h = (partial deriv of V w/resp. to h) = 1/3π r2
dr/dt = rate of change of r (radius: in) with respect to t
(time: s) = 1.8in/s
dh/dt = rate of change of h (height: in) with respect to t
(time: s) = -2.5in/s
è See how the units match? Check them – this will help prevent you from
switching the values up
dV/dr is volume (in3) over radius (in) and so should be in2
(it is: 2/3π r h = 2/3π (in)(in))
dr/dt is radius (in) over time (s) and so should be in/s (it is: 1.8in/s)
Now
the easy part: plugin to the dV/dt = … equation and solve:

è Double check your units one more time – dV/dt =
volume (in3) over time (s) = in3/s (checks)