Download d dx [ √ x − 2 x3 + ex] = d [x1/2]+2 d [x−3] + d [ex] = 1 (b) d dx [ 3x2

Calculus I
Review Material for Exam 2 Solutions
February 16, 2017
Math 151
1. Evaluate the following :
d √
d
d
d
[ x − x23 + ex ] = dx
[x1/2 ] + 2 dx
[x−3 ] + dx
[ex ] = 21 x−1/2 + 6x−4 + ex
dx
¸
·
d 3x2 − 4x − 2
d
d
d
√
= 3 dx
[x3/2 ] − 4 dx
[x1/2 ] − 2 dx
[x−1/2 ] = 29 x1/2 − 2x−1/2 + x−3/2
(b)
dx
x
(a)
(c) y = (y 2 − y12 )(2 + 3y), find y
y = 2y 2 + 3y 3 − 2y −2 − 3y −1
y 0 = 4y + 9y 2 + 4y −3 + 3y −2
√
3
Product
(d) y = 5t t2
Rule
√
dy
dx
dy
dx
0
d
d 2/3
= 5t dx
[t ] + t2 dx
[5t ]
¢
¤
£
¡
2/3 t
t 2 −1/3
= 5 3t
+ t 5 ln(5) = 5t t−1/3 32 + t ln(5)
3
·
¸
3x
d
√
Quotient Rule
(e)
dx 1 + 2 x
¸
·
d
d
(1 + 2x1/2 ) dx
[3x] − 3x dx
[1 + 2x1/2 ]
3x
d
√ =
dx 1 + 2 x
(1 + 2x1/2 )2
1/2
3 + 6x1/2 − 3x1/2
3 + 3x1/2
(1 + 2x ) · 3 − 3x(x−1/2 )
=
=
=
(1 + 2x1/2 )2
(1 + 2x1/2 )2
(1 + 2x1/2 )2
·
¸
xex
d
Quotient Rule
(f)
dx 1 + x2
·
¸
d
d
[xex ] − xex dx
[1 + x2 ]
(1 + x2 ) dx
xex
d
=
dx 1 + x2
+ x2 )2
¤
£ (1
d
d
2
[x] − xex (2x)
(1 + x ) x dx [ex ] + ex dx
=
(1 + x2 )2
(1 + x2 ) [xex + ex ] − 2x2 ex
=
(1 + x2 )2
x
x
xe + e + x3 ex + x2 ex − 2x2 ex
=
(1 + x2 )2
x 3
2
e (x − x + x + 1)
=
(1 + x2 )2
2. Find where the horizontal tangents for the following function occur.
(a) f (x) = 12x − x3
find where f 0 (x) = 0
0
2
f (x) = 12 − 3x → f 0 (x) = 0 → 3(4 − x2 ) = 0 → x = ±2
x
find where y 0 = 0
+9
d
d
(x2 + 9) dx
[x] − x dx
[x2 + 9]
y0 =
(x2 + 9)2
2
x + 9 − x(2x)
y0 =
(x2 + 9)2
9 − x2
(3 + x)(3 − x)
y0 = 2
=
−→ y 0 = 0 when x = ±3
(x + 9)2
(x2 + 9)2
(b) y =
x2
3. Calculate y 0 for the following:
(a) y = ln(sinh(5x))
1
(cosh(5x))(5) = 5 coth(5x)
y0 =
sinh(5x)
sec(2θ)
(b) y =
1 + tan(2θ)
(2 sec(2θ) tan(2θ))(1 + tan(2θ)) − sec(2θ)(sec2 (2θ)(2))
y0 =
(1 + tan(2θ))2
2
2
sec(2θ)[tan(2θ)
+
tan
(2θ) − sec2 (2θ)]
y0 =
(1 + tan(2θ))2
2
sec(2θ)[tan(2θ)
− 1]
y0 =
2
(1 + tan(2θ))
(c) y = x tan−1 (xe2x )
1
[e2x + xe2x (2)]
1 + (xe2x )2
xe2x (1 + 2x)
y 0 = tan−1 (xe2x ) +
1 + x2 e4x
4
2
(d) xy + x y = x + 3y
dy
dy
dy
y 4 + 4xy 3
+ 2xy + x2
=1+3
dx
dx
dx
dy
dy
dy
4xy 3
+ x2
−3
= 1 − y 4 − 2xy
dx
dx
dx
¢
dy ¡
4xy 3 + x2 − 3 = 1 − y 4 − 2xy
dx
1 − y 4 − 2xy
dy
=
dx
4xy 3 + x2 − 3
√
(e) y = arctan(arcsin( x)) Ã
!µ
¶
1
1
1
1
0
√
p
√
y =
= √ √
√
√ 2
1 + (arcsin( x))2
2
x
2 x( 1 − x)(1 + (arcsin( x))2 )
1 − ( x)
y 0 = tan−1 (xe2x ) + x
(f) y = (cos(x))x
ln(y) = ln(cos(x)x )
ln(y) = x ln(cos(x))
1
1 dy
= ln(cos(x)) + x
(− sin(x))
y dx
cos(x)
dy
= y (ln(cos(x)) − x tan(x))
dx
dy
= (cos(x))x (ln(cos(x)) − x tan(x))
dx
√
(g) y = sin(tan(
1 + x2 )) ¢
√
√
¡
0
y = cos tan( 1 + x2 ) [sec2 ( 1 + x2 )][ 12 (1 + x2 )−1/2 ][2x]
√
√
¡
¢
x cos tan( 1 + x2 ) [sec2 ( 1 + x2 )]
0
√
y =
1 + x2
(h) y = tan2 (sin(θ))
y 0 = 2 tan (sin(θ)) [sec2 (sin(θ))][cos(θ)] = 2 cos(θ) tan (sin(θ)) sec2 (sin(θ))
(x2 + 3x)4/3 (3 − x)2/5
x3 (x3 − 3)2
· 2
¸
(x + 3x)4/3 (3 − x)2/5
ln(y) = ln
x3 (x3 − 3)2
2
4
ln(y) = ln(x2 + 3x) + ln(3 − x) − 3 ln(x) − 2 ln(x3 − 3)
3
5
1 dy
4 2x + 3
2 −1
1
3x2
=
+
−3 −2 3
2
y dx ·3 x + 3x 5 3 − x
x
x −3 ¸
dy
4(2x + 3)
2
3
6x2
=y
−
−
−
dx
3(x2 + 3x) 5(3 − x) x x3 − 3
·
¸
dy
(x2 + 3x)4/3 (3 − x)2/5 4(2x + 3)
2
3
6x2
=
−
−
−
dx
x3 (x3 − 3)2
3(x2 + 3x) 5(3 − x) x x3 − 3
(i) y =
4. Find the equation of the tangent line for the curve x2 + 4xy + y 2 = 13 at the point (2, 1).
dy ¯¯
Use the point slope form for the line: y − 1 =
(x − 2)
¯
dx (2,1)
dy
Using implicit differentiation to find
dx
dy
dy
2x + 4y + 4x
+ 2y
=0
dx
dx
dy
(4x + 2y) = −2x − 4y
dx
dy
−2(x + 2y)
−(x + 2y)
=
=
dx
2(2x + y)
(2x + y)
dy ¯¯
−4
−(2 + 2(1))
=
=
¯
dx (2,1)
2(2) + 1
5
Equation of the tangent line is: y − f (a) = f 0 (a)(x − a)
4
y − 1 = − (x − 2)
5
4
13
y =− x+
5
5
5. At what points on the curve y = sin(x) + cos(x), 0 ≤ x ≤ 2π, is the tangent line horizontal?
Need to find x-value(s) where y 0 = 0.
y 0 = cos(x) − sin(x)
y 0 = 0 −→ cos(x) − sin(x) = 0
5π
π
cos(x) = sin(x) −→ x = , x =
4
4
6. At what point on the curve y = [ln(x + 4)]2 is the tangent horizontal?
Need to find x-value
where
y 0 = 0.
·
¸
1
2
ln(x + 4)
y 0 = 2 ln(x + 4)
=
x+4
x+4
2 ln(x + 4)
0
= 0 −→ 2 ln(x + 4) = 0 −→ ln(x + 4) = 0 −→ x = −3
y = 0 −→
x+4
7. A particle moves on a vertical line so that its coordinates at time t is y = t3 − 12t + 3, t ≥ 0.
(a) Find the velocity and acceleration functions.
Velocity function: y 0 (t) = v(t) = 3t2 − 12 = 3(t2 − 4) = 3(t + 2)(t − 2)
Acceleration function: y 00 (t) = a(t) = 6t
(b) When is the particle moving upward and when is it moving downward?
−
t=0
0
+
v(t)
t
t=2
Moving downward from t ∈ [0, 2) while moving upward for t ∈ (2, ∞).
(c) When is the particle speeding up? slowing down?
0
t=0
+
+
a(t)
t
t=2
Slowing down from t ∈ [0, 2) while speeding up for t ∈ (2, ∞).
8. The volume of a right circular cone is V =
height.
πr2 h
, where r is the radius of the base and h is the
3
(a) Find the rate of change of the volume with respect to the height if the radius is constant.
dV
πr2
=
dh
3
(b) Find the rate of change of the volume with respect to the radius if the height is constant.
dV
2πrh
=
dr
3
9. A paper cup has the shape of a cone with height 10 cm and radius 3 cm (at the top). If water is
3
poured into the cup at a rate of 2 cms , how fast is the water level rising when the water is 5 cm
deep?
3
πr2 h
dV
Givens: V =
and
= 2 cms
3
dt
dh
Unknown:
=??? when h = 5cm.
dt
By looking at similar triangles we can find a relationship between r and h.
r
h
3
=
−→ r =
h.
3
10
10
µ
¶2
1
3
3
Now, V = π
h h=
πh3 . Now differentiate with respect to time.
3
10
100
3
9
dh
dh
100 dV
dV
dV
=
πh2
−→
=
−→ substitute h = 5 cm and
= 2 cms .
2
dt
100
dt
dt
9πh dt
dt
dh
100
200 cm
=
(2) =
2
dt
9(5) π
225π s
10. A balloon is rising at a constant speed of 5 ft/s. A boy is cycling along a straight road at a speed
of 15 ft/s. When he passes under the balloon, it is 45 ft. above him. How fast is the distance
between the boy and the balloon increasing 3 sec. later?
dy
= 5 ft/s yields distance upwards y = 5t.
Givens:
dt
dx
= 15 ft/s yields distance right x = 15t.
dt
dz
=??? when t = 3 sec after the boy passes under the balloon.
Unknowns:
dt
y
+
45
z
x
Using the pythagorean theorem we have z 2 = x2 + (y + 45)2 , differentiate implicity with respect
to time.
dz
dx
dy
dz
= 2x
+ 2(y + 45)
which now we want to solve for
2z
dt
dt
dt
dt
dx
dy
x
+ (y + 45)
dz
dt
= dt
. Evaluate at t = 3 where x = 45, y = 15, z = 75
dt
z
45(15) + (60)(5)
975
dz
=
=
= 13 ft/sec.
dt
75
75