# November 10

```Calculus 3
When should we have Test 3?
Monday, November 10
DISCUSS
Critical points &amp; second-derivative test
COMPARE
61(1,3),63(1,2)Tangent planes,normal lines
PRESENT
61(1), 63(1)
ASSIGNMENT Write up 61(3),63(2)
Work 75, 80 critical points
NOTEBOOKS
November 11:
November 18:
Anonymous, Pizza
Resubmissions
Critical points, Extreme points, and 2nd derivative tests
If a surface has a max or min point (hilltop or valley dip),
what is the equation of the tangent plane like there?
What does that tell you about the partials there?
critical values
y = f(x)
a where
f’(a)=0 or
f’(a) undefined
z=f(x,y)
(a,b) where BOTH
fx(a,b)= 0 and
fy(a,b)=0
Example: Find the critical values for z = xy2-6x2-3y2
fx(a,b)= 0
y2-12x = 0
fy(a,b)=0
2xy-6y=0,
so 2y(x-3)=0,
so y=0 or x=3
If y=0, then -12x=0, so x=0.
If x=3, y2-36=0, so y= 6, or -6.
So the three points are (0,0), (3, 6), and (3, -6).
How do you tell if a critical point is a max, min, or
neither? Examples:
Max
Min
Neither
y = -x2 at 0
y = x2 at 0
y = x3 at 0
z = -x2-y2 at (0,0)
z = x2+y2 at (0,0)
z = x2-y2 at (0,0)
For y=f(x), if f’’(a) &gt; 0, then (a, f(a)) is a minimum point.
Why?
What happens if f’’(a) &lt; 0?
What can you conclude if f’’(a)=0?
For y=f(x,y), the second derivative test is given in theorem
78. Let’s use it on our four examples.
```