Notes for MTH 232 exam 2
Prof. Townsend
Fall 2013
Make sure you reread the problem once you think you are done with it to make sure you
answered the question.
Section 24.1 Tangents and Normals.
Given the function y = f ( x ) there are four types of problems you will encounter.
1) Find the equation of the tangent line at the point ( x1 , y1 )
2) Find the equation of the normal line at the point ( x1 , y1 )
3) Find the equation of the tangent line given the slope mtan
4) Find the equation of the normal line given the slope m⊥
To find the equation of the tangent or normal line, you need ( x1 , y1 ) and the slope mtan or m⊥ of
the line. Then solve for the line using either y = mx + b or y = m ( x − x1 ) + y1
In types (1) and (2) find the value of the slope:
dy
Find the slope of the tangent line: mtan =
i.e. take the derivative then plug in x1 and
dx ( x1 ,y1 )
y1 if needed.
In types (3) and (4) find the values of x1 and y1 :
dy
Find the slope of the tangent line: mtan =
. It will be a function of ( x1 , y1 ) . If the
dx ( x1 ,y1 )
problem is type (4), set m⊥ = −
1
. Solve for ( x1 , y1 ) . Find y1 = f ( x1 ) .
mtan
You can check your work by graphing y = f ( x ) . On the TI-89, then use F5/A to draw the
tangent line and give you its equation at ( x1 , y1 ) / To graph the normal line you need to graph
y = mx + b as well then ZoomSqr. Include the tangent line as well since it is so easy to get and
you can then clearly see the perpendicularity of the normal line to the tangent line. To get credit
for the check, make a sketch of what you see.
Section 24.3 Curvilinear Motion
We had two types of problems
(1) Given x(t) and y(t), find the components of the velocity vx =
dx
dy
and vy =
. The
dt
dt
equations you use are on page 702 Equations 24.2->24.5.
(2) Given y(x) and one other piece of information from which you can deduce vx , find vx .
dy dx dy
Then find vy =
= vx .
dx dt dx
Once you have the components of the velocity, find the magnitude and phase.
1) TI command ⎡⎣ vx ,vy ⎤⎦  Polar
Diamond ENTER gives magnitude and phase.
2) SimCalc:
(
Pythagorean theorem gives magnitude: v = vx 2 + vy 2 , atan2 vy ,vx
)
gives the phase. Note the order of the arguments. Be sure to square negative signs inside
Pythagorean theorem.
3) Other: Pythagorean theorem gives magnitude: v = vx 2 + vy 2 . To find the phase find
⎛v ⎞
θ v = tan −1 ⎜ y ⎟ . If vx < 0 then add 180 .
⎝ vx ⎠
Section 24.4 Related Rates
You are given an equation that relates two variables, such as x and y. i.e. y = f ( x ) . You are
typically given one rate. Find the other rate given some value of x. Recall that a rate means a
dy
dx
time derivative. So
and
are related by taking the time derivative of y = f ( x ) , I.e.
dt
dt
dy df ( x ) dx
. Plug in the rate you know, the value of x you know then solve for the other
=
dt
dx dt
rate. Note: If the problem says something like y is increasing at the rate of 3 cm/s, then
dy
dy
= +3 . If the problem says y is decreasing at the rate of 3 cm/s, then
= −3
dt
dt
Section 24.7 Max/Min problems
We had two types of Max/Min problems
1) Find a maximum or minimum for y = f ( x ) .
dy
a) Take the first derivative
dx
dy
b) Set it equal to zero
=0.
dx
⎛ dy
⎞
c) Solve for x. Solve ⎜ = 0, x ⎟
⎝ dx
⎠
d2y
d) Take the second derivative.
dx 2
e) Check its value to guarantee that you found what you were looking for, i.e. the location
of a maximum or a minimum.
d2y
<0
dx 2
d2y
>0
dx 2
d2y
=0
dx 2
Maximum
Minimum
Inflection Point
2) Find a maximum or minimum for y = f ( u,v ) where w = g ( u,v ) . The w = g ( u,v ) function
is the one that has a number in it that nails down the problem. For example, the perimeter=3,
the area=5, the diagonal=16, etc.
a) Solve w = g ( u,v ) for either u or v . For example, choose u .
Solve ( w = g ( u,v ) ,u )
Call that solution uw It is a function of v .
b) Plug that variable into y = f ( u,v ) so now y = f ( uw ,v ) . Since uw is a function of v , you
now have a single variable function to optimize. Note that one of the choices will make
your life a little easier.
c) Since you now have a single variable problem, use the method in (1) above to find the
max/min,