Download My Favorite Applications of Differentiation

Holiday Quiz!
Must Show all Work!
Neatness Counts!
1. Find the equation of the tangent
Line to the graph of y = secxtanx

at the point with x = .
6
Write the answer in exact form!



 2

The point on the graph is ( , sec tan ) = ( , ) and the slope is y’( ),
6
6
6
6 3
6
2
2
3
where y’ = [secxtanx](tanx) + (secx)[sec x] = secxtan x + sec x, so
2
2 3 8 3 10 3



2  3  2 
2 3
8
2 3
8
3
  


m = sec tan2 +sec3 =
=
* 





*

6
6
6
9
9
9
9
9
3  3   3 
3 3
3 3
3
So, with point (

y-
3
10 3
 2
, ) and slope m =
we can find the equation of the tangent line.
6 3
9
10 3
2

=
(x- )
6
3
9
or y =
10 3
5 3 2
10 3
5 3  18

xor y =
x9
27
3
9
27
2. Sketch the graph including all intercepts, asymptotes and extrema:
( x  2)( x  2)
x2  4
y=
=
2
2x  2
2( x  1)( x  1)
We have intercepts:
(-2,0), (2,0) and (0, 2) and
asymptotes y = ½, x = -1, and
x = 1.
3. Same for: y =
3
x  tan x for x in [ 0,2 ] .
3
y’ =
3
3
 sec 2 x
0=
3
3
 sec 2 x 

3
3
 sec 2 x which has no real solns
 there are no critical numbers
Also note that y’ is always positive so
that the
graph must be increasing throughout the interval [ 0,2 ] . The graph also has asymptotes at the sam
values as tanx which are /2 and 3/2.
4. Same for: y = 4x3 – x2.
To find intercepts: 0 = x2( 4x – 1)   (0,0) and (1/4, 0)
To find extrema: y’ = 12x2 – 2x = 2x(6x – 1)   x = 0 and x = 1/6 are the critical numbers.
++++
++++
----0
1/6
By the first derivative test, we have a local maximum at x = 0 and the maximum is 0.
Also, we have a local minimum at x = 1/6 and the local minimum is -0.009
Note that the values are very small and we need to zoom in on the graph to see what is
happening.
5. Find the limits using limit properties:
10 x 3  8 x
7 x 2  10
a) lim
b) lim
x  2 x  3
x  14 x 2  5
Since this is a limit at ∞,
1
We divide both the num. and
=
2
den by the greatest power of x
in the den. which is just x:
10 x 2  8
lim
x  
3
2
x
2
10 x  8
= lim
x 
2
= lim (5x 2  4)
x 
=-∞
x 2  15
c) lim
x 2
x2
 1

* ( x 2  15)
= lim 
x2  x  2

=
-∞ * (4-15) = ∞
6. Two commercial airplanes are flying at 40,000 ft along straight-line courses that intersect at righ
angles. Plane A is approaching the intersection point at a speed of 442 knots (nautical miles per hou
a nautical mile is 2000 yd). Plane B is approaching the intersection at 481 knots. At what rate is the
distance between the planes changing when plane A is 5 nautical miles from the intersection point an
plane B is 12 nautical miles from the intersection point? (Related Rates)
dx
dy
 442knots / hr
&
 481knots / hr
dt
dt
dz
Find:
when the distance from P to boat A is 5 knots and the distance from P to boat B is 12.
dt
Given:
General equation: x2 + y2 = z2
 2 x
dx
dt
 2y
dy
dt
 2z
dz
dt
 x
 (5)( 442)  (12)( 481)  13

dz
dt
dz
dt
dx
dt


y
dy
dt

(5)( 442)
13
z

dz
dt
(12)( 481)
13

  614knots / hr
x
P
z
y
dz
dt
7. A rectangle is to be inscribed under the arch of the curve y = 4cos(0.5x) from x = -  to x =  .
What are the dimensions of the rectangle with the largest area, and what is the largest area?
(Optimization)
The function to be maximized is the area, A.
A = 2x(4cos0.5x) = 8xcos(0.5x)
A’ = 8 cos(0.5x) - 4xsin(0.5x)
To find the critical points we must use Newton’s method on A’, which yields the approximation
formula:
xn+1 = xn -
A' ( x n )
A" ( x n )
= xn -
8 cos(0.5x)  4xsin(0.5x )
 8 sin( 0.5 x)  2 x cos(0.5 x)
To find the first estimate(s) of these critical points we must check the changes in sign of
A’:
A’(0) = 8
A’(  ) = -12 **This means that there is a critical point between 0 and **
A’(2  )
A’(3  )
A’(4  )
A’(5  )
A’(6  )
A’(7  )
A’(8  )
=
=
=
=
=
=
=
-8
+ **This means that there is a critical point between 2 and 3**
+
- **This means that there is a critical point between 4 and 5**
+ **This means that there is a critical point between 6 and 7**
+
This means that we must use Newton’s Method 4 times to get the 4 critical points.
CPT#1
For the first critical point I use 0.5 as my first estimate and then apply the formula (*) to
get the first critical point to be 1.7206.
CPT#2
For the 2nd critical point I use 2.5 as my first estimate and then apply the formula (*) to get
the 2nd critical point to be 6.85.
CPT#3
For the 3rd critical point I use4.5 as my first estimate and then apply the formula (*) to get
the 3rd critical point to be 12.875.
CPT#4
For the 4th critical point I use 6.5 as my first estimate and then apply the formula (*) to get
the 4th critical point to be 19.058.
First Derivative Test:
++++++
------
-------------
++++++++
--
0
1.7206
6.85
12.875
++++++++
A’
19.058
We can see that the local maximums are (1.7206, 8.978), and (192.875, 101.78)
Since this problem is on a closed interval we must also check the endpoint x = 8:
A(8) = 201.06.
After all of this work, the maximum area occurs at x = 8!
8. What are the dimensions of the lightest open-top right circular cylinder can that can hold a
volume of 1000 cm3?
V = 1000 cm3 = r2h 
h=
1000
r 2
Function to be minimized is the Surface Area, S = r2 + 2rh which must be rewritten as a
 1000 
function of 1 variable: S = r2 + 2r  2  = r2 + 2000r-1
 r 
Now we can differentiate S with respect to r:
Dr = 2r – 2000r-2 = 2r-2(r3 – 1000)  We have critical numbers 0 and
------
3
1000
 6.83
r
++++++++
-0
6.83
By the First Derivative Test we have a minimum at r =
3
1000
 6.83 and h =
r
3
1000
 6.83
r

9. Evaluate:

2(3 x  5)8
3
x2
3
 sin
x cos xdx =
2
x
2
x tan xdx =
12. Evaluate:
u
3
2
3
1
x  5 , then du =
3
3 x
(3 x  5) 8
33 x 2
du 
dx
2
6
2
dx  6 u 8 du  u 9  C  (3 x  5) 9  C
9
3
1 4
1
u  C  sin 4 x  C
4
4
x tan xdx , let u = tanx, then du = sec2xdx
1
 udu  2u
 cot
dx , let u =
x cos xdx , let u = sinx, then du = cosxdx
3
 sec
11. Evaluate:
 sec
3
dx = 2 * 3
10.Evaluate:
 sin
2(3 x  5)8
2
xdx =
2
C 
 (csc
2
1
tan 2 x  C
2
x  1)dx   cot x  x  C
13. Find the critical numbers of f(x) = 6 x 3 8  x 2
1
2
1
1
1

1
f’(x) = [18x ] (8  x ) + (6x )[ (8  x 2 ) 2 (2 x) ] = 18x2 (8  x 2 ) 2 -6x4 (8  x 2 ) 2
2
*which we can now factor*
2
2
=6x (8  x )
=
2

2
1
2
3
2
(3( (8  x )  x ) = 6x (8  x )
2
2
2

1
2
2
2
(-4x +24) = -24 x (8  x )
2

1
2
(x2 – 6)
 24 x 2 ( x 2  6)
(8  x )
2
1
2
So that we get critical numbers x = 0, x =  6 , and x =  2 2 with the last 2 being of type II.
14. Find the critical numbers of f(x) =

f’(x) =
=
1
2
1
2
8 x
x2  4
1
2

1
2
1
2
[4 x ]( x  4)  (8 x )[ 2 x] x [4 x ]( x  4)  (8 x )[ 2 x]
 1*

( x 2  4) 2
( x 2  4) 2
2
x
2
[4]( x 2  4)  (8 x )[ 2 x]
1
2
x ( x  4)
2
2

4 x 2  16  16 x 2
1
2
x ( x  4)
2

2
 12 x 2  16
1
2
x ( x  4)
2
2
2

 4(3x 2  4)
1
2
x ( x 2  4) 2
From the derivative it appears that we have cn’s 0 and  2 .
Upon inspection we see that  2 are not in the domain so that the only cn is x = 0 which is
of type II.
15. Evaluate:

sec x  tan x
dx =
cos x
 (sec
2
x  sec x tan x)dx = tanx + secx + C
16. Find the volume of the solid of revolution obtained by revolving the region enclosed by the graph
of y = 2x2 – 2 and y = 3x+3 about the line x = 3.
First let us find the points of intersection: 2x2 – 2 = 3x + 3 
2x2 – 3x – 5 = 0 
(2x – 5)(x + 1) = 0  x = 5/2 and x = -1.
It is easiest to use
shells:
x=3
cylindrica
x
The radius of each cylindrical shell is r = 3 – x the height is the distance between the two
graphs, h = (3x+3) – (2x2 – 2) = -2x2 + 3x + 5.
2.5

V = 2  (3  x)( 2 x 2  3 x  5)dx = . . .  202.04 cubic units
1
17. Same region as #16 revolved about the line y = -4.
Here it is best to use washers:
Ro is the distance between the the axis of revolution and farthest function bounding the region
(the line):
Ro = (3x + 3) – (- 4) = 3x + 7
RI is the distance between the axis of revolution and the closest function bounding the region
(the parabola):
RI = (2x2 – 2) – (-4) = 2x2 + 2

V=
2.5
 [(3x  7)
2
 (2 x 2  2) 2 ]dx = . . .  610.62 cubic units
1
Ro
& RI