3 y 2 y ' y ' x 2  2 xy  2 x  6 yy '  0
APCalcBC1 HW#38
A214
Find f '( x)
23
f ( x)  cos 5 x
2 xy  2 x
00
0
(0,3) 
2
2
3y  x  6 y
27  18
Observe that the curve has a
horizontal tangent at (0,3).
A222
Let V be the volume of a cylinder having
7
height h and radius r, and assume that h and r vary
with time.
(a) How are dV/dt, dh/dt, and dr/dt related?
V   r 2h
y' 
1
1
5sin(5 x)
f '( x)  cos 2 (5 x)[ sin(5 x)](5)  
2
2 cos(5 x)
A276
Find the open intervals on which f is concave
18c
up.
x
x2  2  2 x2
 x2  2
f ( x)  2
, f '( x) 

x 2
( x 2  2)2
( x 2  2)2
2 x( x 2  2) 2  2( x 2  2)(2 x)( x 2  2)
f ''( x) 
( x 2  2) 4
dV
dh 
 dr
   2r h  r 2 
dt
dt 
 dt
(b) At a certain instant, the height is 6 in
and increasing at 1 in/s, while the
radius is 10 in and decreasing at 1 in/s. How fast is the
volume changing at that instant? Is the volume
increasing or decreasing at that instant?
2 x( x 2  2)[( x 2  2)  2( x 2  2)]

( x 2  2) 4

2 x( x 2  6)
 0  2 x( x 2  6)  0
( x 2  2)3
 2 x( x  6)( x  6)  0
 x  0, x  6, x   6
Note: 6  2.4
f ''( x)  2 x( x  6)( x  6)
0
+
0
0
+




 6
6
0
 f is concave up on ( 6, 0)  ( 6, )
below
For t  0 , the position of a particle is given
8
by s(t )  t 5  3t 4  3t  1 . Find the acceleration when
the particle is at rest.
v(t )  5t 4  12t 3  3  0
Use G.C. X[0,3], Y[-30,10], ZERO
t  2.441
a(t )  20t 3  36t 2
a(2.441)  76.3877
A242
Use implicit differentiation to find the slope
30
of the tangent line to the curve at the specified point
and check that your answer is consistent with the
accompanying graph.
y3  yx 2  x 2  3 y 2  0, (0,3) (trisectrix)
dV
  (2(10)(1)(6)  1 102 )  20 in3 / s
dt
The volume is decreasing at 20 in3 / s .
A223
A conical water tank with vertex down has a radius of
25
10 ft at the top and is 24 ft high. If water flows into the tank at a
3
rate of 20 ft / min , how fast is the depth of the water
increasing when the water is 16 ft
dV
 20 ft 3 / min
dt
dh
deep?
 ? when h=16 ft
dt
1
V=  r 2 h
3
r
h

Using similar triangles,
, 10 h = 24r
10 24
10h 5h
r

. Substitute in the formula to get
24 12
2
1  5h 
1 25h3
V    h 
3  12 
3 144
2
dV 25 h dh
25 (16) 2 dh

 20 
dt
144 dt
144 dt
dh
9
 
ft / min . The depth of the water is
dt 20
9
ft / min
increasing at the rate of
20
F 373
6c
Open Box II (Project): For this project you
are to investigate the volume of the box formed by
HW 38
cutting out squares from the four corners of a 20 by
12 unit piece of graph paper and folding up the edges
to form a box without a top.
Find the value of x that gives the maximum volume.
What is this volume?
2nd option:
Use GC to find the maximum volume directly by
using the volume function:
V  4 x3  64 x 2  240 x
Y1:= 4 x ^ 3  64 x 2  240 x
Windows: X[0,6]. Y[0, 300]
Press Graph to see:
2nd , CALC, Maximum,
Left 1 space, ENTER,
Right 2 spaces, ENTER,
Left 1 space, ENTER
Answer:
V  (20  2 x)(12  2 x) x
(20  2 x)(12  2 x) x  0  x  0, x  10, x  6
To find domain of V, take min(6,10) = 6
0 x6
V  (20  2 x)(12  2 x) x
V  4 x  64 x  240 x
3
2
V '  12 x 2  128 x  240
12 x 2  128 x  240  0
3 x  32 x  60  0
2
Use G.C.
Y1:= 3X 2  32 X  60
Set: WINDOW:
X[0, 6] domain of V
Y[-30,10]
GRAPH
Use ZERO to find the zero of the derivative function
in the valid interval. TRACE, move cursor to
approximate the zero.
2nd CALC, ZERO, Left 1 space, ENTER to accept
the left bound.
Right 2 spaces, ENTER to accept the right bound.
Left 1 space, ENTER to accept the Guess number.
X=2.427400….
Since the derivative, V ' changes from positive to
negative, V is a maximum at this value of x.
At x  2.4274
V  4(2.4274)3  64(2.4274)2  240  2.4274
 262.682
A334
8
262.6823
x=2.4274
Verify that the hypotheses of the MVT are
satisfied on the given interval, and find all values of c
in that interval that satisfy the conclusion of the
theorem.
f ( x)  x3  x  4; [1, 2]
Since f is a polynomial function, it is differentiable
everywhere.
f (2)  f (1) 6  (6) 12


4
2  (1)
3
3
f '( x)  3x 2  1  4
x  1 reject -1
 c  1  (1, 2) .