Download Lesson Plan #6

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
1
Lesson Plan #31
Date: Tuesday November 30th, 2010
Class: AP Calculus
Topic: Related Rates involving trigonometric functions.
Aim: How do we solve Related Rates questions involving trigonometric functions?
Objectives:
1) Students will be able to solve Related Rates questions involving trigonometric functions
HW# 31:
Page 240#’s 34, 35
Do Now:
1) Given A 
1
dA
ab sin  , find
dt
2
Answer
2) For Do Now 1), evaluate
dA
da
db
d
 2,
 7,
 0 and   30o
when a  5, b  12,
dt
dt
dt
dt
Answer
3) In words, describe a problem that you are solving with Do Now examples 1 and 2.
Procedure:
Write the Aim and Do Now
Get students working!
Take attendance
Give back work
Go over the HW
Collect HW
Go over the Do Now
When we are finding rates of change of two or more variables that are changing with respect to time, these rates are called Related
Rates. Today, let’s see how we find Related Rates involving trig functions.
2
1) A hot-air balloon, rising straight up from a level field, is tracked by a range finder 500 ft from the lift-off point. At the
moment the range finder’s angle of elevation is

, the angle is increasing at a rate of 0.14 radians/minute.
4
How fast is
the balloon rising?
y
Range Finder

500 feet
Ans.
2) Find the rate of change in the angle of elevation of the camera shown in the figure 10 seconds after liftoff, if the height of
rocket is given by the position function s  50t
2
s

2000 ft.
Ans.
3
Sample Test Questions:
20 x 2  13x  5
x
5  4 x3
B) 
C) 0
D) 5
1) Evaluate
A) -5
2)
(2 x  1)(3  x)
x ( x  1)( x  3)
3) If
D)
B) -2
C) 2

6 x( x 2  2) 2
x
D) 3
E) nonexistent

3
f ( x)  ( x  1) x 2  2 , then f ' ( x) 
2

2
B)
6 x( x  1)( x 2  2) 2
 2 7 x  6x  2
4) If f ( x)
A) -2cos3
2

2
B) -2sin3cos3
E)
x
2


2
 2 x 2  3x  1
 3( x  1)( x  2)
2
2
B)
C) 6cos3
D) 2sin3cos3
E) 6sin3cos3
1
2
, then f '   equals
x
 
2
D) 
C) -1

6) If sin( xy)  y , then
A) sec(xy)

C)
 sin 2 (3  x), then f ' (0) =
5) If f ( x)  x cos
A)
E) 1
lim
A) -3
A)
lim
dy
equals
dx
B) ycos(xy)-1
C)

2
1  y cos( xy)
x cos( xy)
E) 1
D)
y cos( xy)
1  x cos( xy)
E) cos(xy)
Related documents