David Underhill – Calculus 3 w/Optimization (Putt)
Test 2 Notes
Functions of Several Variables – Section 11.1, P.756
 z = f(x, y) = <expression containing x and y>
 Level Curve – hold a variable in a function constant and then graph it
o Typically done when there are four dimensions ( f(x, y, z) = … ) because you cannot graph 4D
o Each variable held constant reduces the number of dimensions by 1
o Example (from quiz 1): k = f(x, y) = x2 + y; k = 2 … solve for y= and graph
x 2  y  2  parabola
y  x2  2
Partial Derivatives – Section 11.3, P.776
 Partial derivative – holds all but one var constant and take deriv with respect to the var not being held constant
Definition: dz  f x ( x, y )  g ' (a) where g (a)  f (a, b)



dx
Important to understand that fx is the derivative of z (because z = f(x, y)) with respect to x
Notation: Derivative of A with respect to B means dA/dB – important to understand for chain rule later
Notation: if z = f(x, y) then fx(x, y) means dz/dx
o

Example (from quiz 2):
f ( x, y, z )  x 2  y 4  x 2 z 2
f x ( x, y, z )  2 x  0  2 x( z 2 )
fxy means take the partial derivative with respect to x to get an answer. Then, take the partial derivative of that
answer with respect to y
o Example (multiple partial derivatives): Find fxy(x, y):
f ( x, y , z )  4 x 4 y 2  4 x 2  y 5
First, find the d/dx : f x ( x, y, z )  16 x 3 y 2  8 x  0
Next, find the d/dy (from last step' s ans) : f xy ( x, y, z )  32 x 3 y  0  0

Clairant’s Theorem: f xy ( x, y, z...)  f yx ( x, y, z...)

Finding partial derivatives from a chart
o Process (for a normal x by y type chart [z = f(x, y))
 Hold one variable constant (lets make y constant first)
 Pick a y value close to the number you’re trying to approximate; let this value = b
 Pick a value just above b (call it a) and just below b (call it c) (still holding y constant)
 So, the equation for the partial deriv of x (dz/dx) is:
f x ( x, y ) 
o
1  f (a, y )  f (b, y ) f (b, y )  f (c, y ) 


  some value (plugin x, a, b, and c)
2
ab
bc

 Do same process to find dz/dy
Example *: Find the rate of change of x and y near y=25 and x=30 from a chart
Chart
x=10
x=20
x=30
x=40
x=50
y=15
4
7
13
21
29
y=20
5
8
16
25
36
y=25
5
9
18
31
40
y=30
5
9
21
31
45
1) Hold y constant at y=25; let b = 30 so: a = 40, c = 20 … find dz/dx:
f x ( x, y ) 
1  f (40,25)  f (30,25) f (30,25)  f (20,25)  1  31  18 18  9 



 
  1.1
2
40  30
30  20
10 
 2  10
2) Hold x constant at x=30; let b = 25 so: a = 30, c = 20 … find dz/dy:
f y ( x, y ) 
1  f (30,30)  f (30,25) f (30,25)  f (30,20)  1  21  18 18  16 



 
  .5
2
30  25
25  20
5 
 2 5
1
Tangent Planes and Linearization – Section 11.4, P.788
 Tangent Plane Equation
z  z 0  f x ( x0 , y 0 )( x  x0 )  f y ( x0 , y 0 )( y  y 0 )
o
Example (from quiz 2):
z  y ln x; P(1, 4, 0)  (x 0 , y 0 , z 0 )
y 4
  4; f y ( x0 , y 0 )  ln x  ln( 1)  0
x 1
Plugin val ues to tangent plane eq : z  0  (4)( x  1)  (0)( y  4)
Find partial derivs and plugin val ues : f x ( x0 , y 0 ) 
Simplify : z  4 x  4

Linearization – used to approximate values
L( x, y )  f ( x0 , y 0 )  f x ( x0 , y 0 )( x  x0 )  f y ( x0 , y 0 )( y  y 0 )
o
o
o
Use: find the partial derivatives, then you can plugin values and get an approximate of x0, y0
Example
Example: finding Linearization from a chart (continued from Example * above) … find L(32, 23)
Find Linearization at base point (let x0 and y0 equal the middle value b from above … so x0 = 30; y0 = 25)
L( x, y )  f (30,25)  f x (30,25)( x  30)  f y (30,25)( y  25)
L( x, y )  18  (1.1)( x  30)  (.5)( y  25)
Plugin the x and y to find the approximate value at x, y:
L(32,23)  18  (1.1)(32  30)  (.5)( 23  25)  19.2
Chain Rule – Section 11.5, P.796
dz dz dx dz dy



 ...
dt dx dt dy dt
o Variable Types: z (dependent); x, y (intermediate); t (independent)
Fx
Fy

Implicit Differentiation: y '  

Problem Types (3):
o 1) Find z’ given: z = f(x, y) = <an expression with x and y); x = <exp. with t>; y = <exp. with t>
 Ex1: z = x2 + xy + y2; x = 5 + 2t2; y = 5t
Find the pieces you need to get dz/dt  dz/dx, dx/dt, dz/dy, and dy/dt:
dx
dz
 2x  y  0
 0  2t
dx
dt
dz
 0  x  2y
dy
dy
5
dt
dz dz dx dz dy


 (2 x  y )( 2t )  ( x  2 y )(5)
dt dx dt dy dt
o
2) Find z’ given: z = f(x, y) … x = g(w) = p; g’(p) = a … y = h(w) = q; h’(q) = b … fx(p, q) = c; fy(p, q) = d
dz dz dx dz dy


 (c)(a)  (d )(b)
dt dx dt dy dt
 Ex2:Find z’
Given : z  f ( g (t ), h(t ))... g (3)  2; g ' (3)  5...h(3)  7...h' (3)  4... f x (2,7)  6... f y (2,7)  8
dz dz dx dz dy


 (6)(5)  (8)( 4)  62
dt dx dt dy dt
2
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)
 Ex3: You are working with a cylinder (V=1/3πr2h). The radius is increasing at 1.8in/s and the
height is decreasing at 2.5in/s. At what rate is volume changing when r=120in and h=140in?
The equations : V  f(r, h)  1/3 r 2 h and
dV dV dr dV dh


dt
dr dt dh dt
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
 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:
dV dV dr dV dh  2 
 



rh 1.8in / s    r 2  2.5in / s 
dt
dr dt dh dt  3

3 
dV  2




(120in )(140in ) 1.8in / s    (120in ) 2  2.5in / s 
dt  3

3

dV
 63334in 3 / s  37699in 3 / s  25635in 3 / s
dt
 Double check your units one more time – dV/dt = volume (in3) over time (s) = in3/s (checks)
o
Volumes:
 Cylinder / Right-cone: V = 1/3πr2h
 Ellipsoid: V = 4/3πr1r2r3
 Pyramid: V = 1/3bh
 Sphere: V = 4/3πr3
3