Math 20A. Summer Session I 2016. Solutions to Problems similar to
Homework 4.
3.6 Trig derivatives
#1. Compute the derivative of y = 3 csc x − ex cos x. Ans:
y 0 = −3 csc x cot x + ex (sin x − cos x) .
#2. Find the equation of the tangent line of y = tan x + cot x at x =
Ans:
y 0 = sec2 x − csc2 x.
y 0 π3 = sec2 π3 − csc2 π3 = 38 . Eqn:
π
3.
4
8
π
y− √ =
x−
.
3
3
3
#3. Show that both y = sin(3x) and y = cos(3x) satisfy y 00 = −9y. You
may use the Chain Rule. Ans: If y = sin(3x), then y 0 = 3 cos(3x) and y 00 =
−9 sin(3x) = −9y. Similarly, ff y = cos(3x), then y 0 = −3 sin(3x) and y 00 =
−9 cos(3x) = −9y.
#4. Find y (163) , where y = sin(3x). You may use the Chain Rule. Ans:
Since 163 = 40 · 4 + 3, we have
y (163) = −3163 cos(3x).
3.7 Chain Rule
#5. Compute the derivative of y = sin t3 e4t . Ans: y 0 = e4t 4t3 + 3t2 cos t3 e4t .
#6. Compute the derivatives of f (g (x)) and g (f (x)), where f (x) = cot x
and g (x) = e3x . Ans:
d
d
f (g (x)) =
cot e3x = −3e3x csc2 e3x
dx
dx
d 3 cot x
d
g (f (x)) =
e
= −3 csc2 x e3 cot x .
dx
dx
2
d
x
cos x
) . Ans:
#7. Compute the higher derivative dx
2 (cos(e ) + e
ecos x − cos2 x − cos x + 1 − e2x cos (ex ) − ex sin (ex ) .
#8. Compute the second derivative of of sin (g (x)) at x = 1 assuming that
g (1) = π6 , g 0 (1) = 2 and g 00 (1) = −3. Ans:
d
sin (g (x)) = g 0 (x) cos (g (x)) ,
dx
d2
2
sin (g (x)) = g 00 (x) cos (g (x)) − (g 0 (x)) sin (g (x)) ,
dx2
1
so
d2
2
sin (g (x)) |x=1 = g 00 (1) cos (g (1)) − (g 0 (1)) sin (g (1))
dx2
π
π
= −3 cos
− 22 sin
6
√ 6
3 3
=−
− 2.
2
3.8 Implicit Differentiation
#9. Calculate the derivative with respect to x of x3 + y 10 = 126 at the
point (5, 1) . Ans:
Using implicit differentiation,
dy
=0
dx
dy
3x2
=⇒
=−
dx
10y 9
3x2 + 10y 9 ·
dy =⇒
dx =−
(x,y)=(5,1)
15
3(5)2
= −
9
10(1)
2
#10. Find the derivative of y = cos−1 x−1 + csc−1 x. Ans:
By using trig laws,
dy
1
= q
dx
x2 1 −
−
1
x2
1
1
1
√
= √
− √
2
2
|x| x − 1
x x − 1 |x| x2 − 1
#11. Show that (cot−1 x)0 = − sin2 (cot−1 x). Then using the Pythagorean
theorem to prove that (cot−1 x)0 = − x21+1 Ans:
y = cot−1 x
=⇒ cot(y) = cot(cot−1 x) = x
dy
= 1 (by implicit differentiation)
=⇒ − csc2 (y)
dx
dy
=⇒ (cot−1 x)0 = y 0 =
= − sin2 (y) = − sin2 (cot−1 x)
dx
Next, note that cot−1 x, by definition of cotangent, corresponds to some
angle y where the cotangent of that angle is x1 . So we can have a triangle
with an adjacent leg of length x and an opposite leg of length √
1, and it would
correspond to that angle y. The hypotenuse of that triangle is x2 + 1 by the
Pythagorean theorem. This means that the sin of that angle is √x21+1 and
therefore:
(cot
−1
0
2
x) = − sin (cot
−1
2
1
1
√
x) = −
=− 2
2
x
+1
x +1
2
#12. Find an equation for the tangent line of x2 y + xy 4 = 6 at the point
(2, 1) . Ans:
Using implicit differentiation:
dy
dy
+ y 4 + 4xy 3 ·
=0
dx
dx
dy
2xy + y 4
=⇒
=− 2
dx
x + 4xy 3
2(2)(1) + (1)4
dy 5
=− 2
=⇒
= −
dx (2) + 4(2)(1)3
12
2xy + x2 ·
(1)
(2)
(3)
(x,y)=(2,1)
3.9 Exponentials and Logarithms
2 x
#13. Calculate the derivative of y = ln x−1
. Ans:
y = 2 ln(x) − ln(x − 1)
y0 =
1
2
−
x x−1
#14. Find an equation for the tangent line of y = x + ln (sec x) at x =
Ans:
7π
4 .
y 0 = 1 + tan x
7π
=
1
+
tan
=1−1=0
7π
4
x= 4
7π
7π
y 7π =
+ ln sec
4
4
x= 4
√ 7π
+ ln 2
=
4
Equation of line:
y0 √ 7π
y = 0x + b =⇒ y = b =⇒ b =
+ ln 2
4
√ 7π
+ ln 2 is the equation of the line
=⇒ y =
4
#15. By simplifying first, calculate the derivatives of (a) y =
2
5x3 e(ln x)
y=
.
x2+ln x
Ans:
a) y 0 = 0
b) y 0 = 5
3
eln 2x
, (b)
x
(HINT: Use the log/exponential laws covered in week 1)
2
#16. Use logarithmic differentiation to find the derivative of y = xx (that
2
is, y = x(x ) ). Ans:
2
ln y = ln xx
ln y = x2 ln(x)
2
dy
= xx (2x ln x + x)
dx
3.10 Related Rates
#17. The radius of a sphere is expanding at a rate of 5 cm/s. Find the
rate of change of the volume of the sphere with respect to time when r = 2 cm.
Ans:
4 3
πr
3
dV
dr
= 4πr2
dt
dt
V =
dV
dt
= 80π cm/s
r=2, dr
dt =5
#18. A 5-m ladder is sliding down a wall. Suppose the bottom is 1 m from
the wall at t = 0 s. Assuming that the bottom slides away from the wall at a
rate of 0.5 m/s, find the velocity of the top of the ladder at t = 4 s. Ans:
At t = 4 seconds, the ladder would have traveled another 0.5(4) = 2 meters
from the wall, so it is 3 m from the wall at that
√ time. We use the Pythagorean
theorem to find that the top of the ladder is 52 − 32 = 4 meters from the floor.
Then we use the Pythagorean theorem to solve the related rates problem. We
let x represent the distance of the bottom of the ladder from the wall, and y
represent the distance of the top of the ladder from the floor.
x2 + y 2 = L2 = 52 = 25
dx
dy
+ 2y ·
=0
2x ·
dt
dt
dy
x dx
=− ·
dt
y dt
dy 3
3
= − (0.5) = − m/s
dt
4
8
dx
x=3,y=4, dt =0.5
#19. See Figure 8 on p. 186 with different values. A road perpendicular
to a highway leads to a pokemon gym located 5 km away. A motorcycle travels
4
past the pokemon gym at a speed of 50 km/h. How fast is the distance between
the motorcycle and the pokemon gym increasing when the motorcycle is 12 km
past the intersection of the highway and the road? Ans:
√
We use the Pythagorean theorem to find that the motorcycle is 52 + 122 =
13 km from the pokemon gym. Then we use the Pythagorean theorem to solve
the related rates problem. We let x represent the distance from the motorcycle
to the intersection, and y represent the distance from the motorcycle to the
pokemon gym.
25 + x2 = y 2
dy
dx
= 2y ·
2x ·
dt
dt
dy
x dx
= ·
dt
y dt
dy dt dx
x=12,y=13, dt =50
=
12
600
(50) =
km/h
13
13
#20. See Figure 9 on p. 186 with different values. A woman of height 1.5
m walks away from a 4.5-m lamppost at a speed of 1.2 m/s. Find the rate at
which her shadow is increasing in length. Ans:
We use proportions to relate the rates. Let x be the distance from the
lamppost to the woman, and let y be the length of the woman’s shadow. The
distance from the lamppost to the end of the woman’s shadow is x + y.
4.5
x+y
1
=3=
=⇒ 3y = x + y =⇒ y = x
1.5
y
2
1 dx
dy
= ·
dt 2 dt
dy 1
= (1.2) = 0.6 m/s
dt dx
2
dt
=1.2
4.1 Linear Approximation
#21. Estimate 36.011/2 − 361/2 using the linear approximation and find the
error using a calculator.
Ans:
√
Let f (x) = x − 6. Then f 0 (x) = 2√1x . Using linear approximation, we find
the linearization L(x)
5
L(x) = f 0 (36)(x − 36) + f (36) =
1
(x − 36) + 0
12
1
1
(36.01 − 36) =
≈ 0.0008333
12
1200
|L(36) − f (36)|
≈ 0.00006944
Error =
|f (36)|
L(36.01) =
√
#22. Find the linearization of y = e
x−1
at x = 16. Ans:
y|x=9 = e3
√
0
y =
e
x−1
√
2 x
e3
8
L(x) = f 0 (16)(x − 16) + f (16)
y 0 |x=16 =
L(x) =
e3
(x − 16) + e3
8
#23. Approximate tan (0.001) using linearization. Then use a calculator to
compute the percentage error. Ans:
y = tan x
y|x=0 = 0
y 0 = sec2 x
y 0 |x=0 = 1
L(x) = f 0 (0)(x − 0) + f (0)
L(x) = x =⇒ L(0.001) = 0.001
Error =
|L(0.001) − f (0.001)|
≈ 3.333 × 10−7
|f (0.001)|
1/3
#24. Approximate (125.01)
using linearization. Then use a calculator
to compute the percentage error. Ans:
6
y=
√
3
x
y|x=125 = 5
1
y0 = √
3
3 x2
1
y 0 |x=5 =
75
L(x) = f 0 (125)(x − 125) + f (125)
1
(x − 125) + 5 =⇒ L(125.01) ≈ 5.0001333
L(x) =
75
|L(125.01) − f (125.01)|
≈ 7.111 × 10−10 × 10−7
Error =
|f (125.01)|
4.2 Extreme Values
#25. Find all critical points of the function y = x2 + x + 21 e2x . Ans:
y 0 = e2x (2x + 1) + (2x2 + 2x + 1)e2x = 2e2x (x + 1)2
Critical points at x = −1
#26. Do Exercise 21 on p. 207, but with f (x) = x2 − 6x + 5. Ans:
a) f 0 (x) = 2x − 6 so critical point at x = 3. f (3) = −4
b) f (0) = 5 and f (4) = −3
c) Minimum is −4 at x = 3, maximum is 5 at x = 0
d) f (1) = 0. There are no critical points on the interval [0, 1], so we only
compare endpoints. Minimum is 0 at x = 1, maximum is 5 at x = 0.
#27. By comparing critical points and endpoints, find the minimum and
maximum value of the function y = 5x − 6ex + 12 e2x on the interval [−1, 6] .
Ans:
y 0 = 5 − 6ex + e2x = (ex − 1)(ex − 5)
Critical points at x = 0, ln 5
yx=0 = −5.5
yx=ln 5 ≈ −9.4528 =⇒ minimum value
yx=−1 ≈ −7.1396
yx=6 ≈ 78986.822 =⇒ maximum value
#28. By comparing critical points and endpoints, find the minimum and
maximum value of the function y = sin 2x − cos 2x on the interval [0, 2π] . Ans:
7
y 0 = 2 cos 2x + 2 sin 2x
3π 7π 11π 15π
,
,
,
8 8
8
8
≈ 1.4142 =⇒ maximum value
Critical points at x =
yx= 3π
8
yx= 7π
≈ −1.4142 =⇒ minimum value
8
≈ 1.4142 =⇒ maximum value
yx= 11π
8
yx= 15π
≈ −1.4142 =⇒ minimum value
8
yx=0 = −1
yx=2π = −1
8