1 Solve Problems about Composite...
Transcript of 1 Solve Problems about Composite...
1 Solve Problems about Composite Functions1. Differentiate each of the following with respect to x:
(a) f (x) = sin 2x
Solution:
f = sin u u = 2x⇒ f ′ (u) = cos u ⇒ u′ (x) = 2
⇒ f ′ (x) = f ′ (u) × u′ (x) = cos u × 2 = 2 cos 2x
(b) f (x) = cos 2x
Solution:
f = cos u u = 2x⇒ f ′ (u) = − sin u ⇒ u′ (x) = 2
⇒ f ′ (x) = f ′ (u) × u′ (x) = − sin u × 2 = −2 sin 2x
(c) f (x) = tan 2x
Solution:
f = tan u u = 2x
⇒ f ′ (u) =1
cos2 u⇒ u′ (x) = 2
⇒ f ′ (x) = f ′ (u) × u′ (x) =1
cos2 u× 2 =
2cos2 2x
Solutions - Differentiation with Trig Page 1 of 26
(d) y = sin2 x
Solution:
y = u2 u = sin x
⇒dydu
= 2u ⇒dudx
= cos x
⇒dydx
=dydu×
dudx
= 2u × cos x = 2 sin x cos x
(e) y = cos2 x
Solution:
y = u2 u = cos x
⇒dydu
= 2u ⇒dudx
= − sin x
⇒dydx
=dydu×
dudx
= 2u × − sin x = −2 cos x sin x
(f) y = tan2 x
Solution:
y = u2 u = tan x
⇒dydu
= 2u ⇒dudx
=1
cos2 x
⇒dydx
=dydu×
dudx
= 2u ×1
cos2 x=
2 tan xcos2 x
Solutions - Differentiation with Trig Page 2 of 26
(g) y = sin3 (2x)
Solution:
y = u3 u = sin v v = 2x
⇒dydu
= 3u2 ⇒dudv
= cos v ⇒dvdx
= 2
⇒dydx
=dydu×
dudv×
dvdx
= 3u2 × cos v × 2
= 6 sin2 (2x) cos (2x)
(h) f (x) = cos2(x2 + 7
)Solution:
f = u2 u = cos v = x2 + 7⇒ f ′ (u) = 2u ⇒ u′ (v) = − sin v ⇒ v′x = 2x
⇒ f ′ (x) = f ′ (u)× ⇒ u′ (v) × v′ (x)= 2u × − sin v × 2x
= −4x cos(x2 + 7
)sin
(x2 + 7
)
(i) y = tan4(2x3 + 4x2 − 6x − 8
)Solution:
y = u4 u = tan v v = 2x3 + 4x2 − 6x − 8
⇒dydu
= 4u3 ⇒dudv
=1
cos2 v⇒
dvdx
= 6x2 + 8x − 6
⇒dydx
=dydu×
dudv×
dvdx
= 4u3 ×1
cos2 v×
(6x2 + 8x − 6
)=
4(6x2 + 8x − 6
)tan3
(2x3 + 4x2 − 6x − 8
)cos2 (
2x3 + 4x2 − 6x − 8)
Solutions - Differentiation with Trig Page 3 of 26
2 Solve Problems about Products
2. Differentiate each of the following with respect to x:
(a) f (x) = x sin x
Solution:
u = x v = sin x
⇒dudx
= 1 ⇒dvdx
= cos x
⇒ f ′ (x) = x cos x + sin x × 1= x cos x + sin x
(b) f (x) = x2 cos x
Solution:
u = x2 v = cos x
⇒dudx
= 2x ⇒dvdx
= − sin x
⇒ f ′ (x) = x2 × − sin x + cos x × 2x
= −x2 sin x + 2x cos x
(c) f (x) = ex tan x
Solution:
u = ex v = tan x
⇒dudx
= ex ⇒dvdx
=1
cos2 x
⇒ f ′ (x) = ex ×1
cos2 x+ tan x × ex
=ex
cos2 x+ ex tan x
Solutions - Differentiation with Trig Page 4 of 26
(d) f (x) = ln x sin x
Solution:
u = ln x v = sin x
⇒dudx
=1x
⇒dvdx
= cos x
⇒ f ′ (x) = ln x × cos x + sin x ×1x
= ln x cos x +sin x
x
(e) f (x) =(x2 + 4x + 3
)sin x
Solution:
u = x2 + 4x + 3 v = sin x
⇒dudx
= 2x + 4 ⇒dvdx
= cos x
⇒ f ′ (x) =(x2 + 4x + 3
)× cos x + sin x × (2x + 4)
=(x2 + 4x + 3
)cos x + (2x + 4) sin x
(f) y = 2 cos x sin x
Solution:
u = 2 cos x v = sin x
⇒dudx
= −2 sin x ⇒dvdx
= cos x
⇒dydx
= 2 cos x × cos x + sin x × −2 sin x
= 2 cos2 x − 2 sin x
Solutions - Differentiation with Trig Page 5 of 26
(g) y = ex ln x sin x
Solution: This is a triple product, but the product rule only deals with two. Thus, weneed to apply the product rule twice. Only the product with sin x is shown here. Trydifferentiating ex ln x yourself and see if you get the answer shown below.
u = ex ln x v = sin x
⇒dudx
=ex
x+ ex ln x ⇒
dvdx
= cos x
⇒dydx
= ex ln x × cos x + sin x ×(ex
x+ ex ln x
)= ex ln x cos x + ex ln x sin x +
ex sin xx
3 Solve Problems about Quotients
3. Differentiate each of the following with respect to x:
(a) f (x) =sin x
x
Solution:
u = sin x v = x
⇒dudx
= cos x ⇒dvdx
= 1
⇒ f ′ (x) =x cos x − 1 sin x
x2
=x cos x − sin x
x2
Solutions - Differentiation with Trig Page 6 of 26
(b) f (x) =1
tan x
Solution:
u = 1 v = tan x
⇒dudx
= 0 ⇒dvdx
=1
cos2 x
⇒ f ′ (x) =tan x × 0 − 1
cos2 x
tan2 x
=1
tan2 x cos2 x
(c) f (x) =x
cos x
Solution:
u = x v = cos x
⇒dudx
= 1 ⇒dvdx
= − sin x
⇒ f ′ (x) =cos x × 1 − x × − sin x
cos2 x
=cos x + x sin x
cos2 x
Solutions - Differentiation with Trig Page 7 of 26
(d) f (x) =sin x
1 + cos x
Solution:
u = sin x v = 1 + cos x
⇒dudx
= cos x ⇒dvdx
= − sin x
⇒ f ′ (x) =(1 + cos x) × cos x − sin x × − sin x
(1 + cos x)2
=cos x + cos2 x + sin2 x
(1 + cos x)2
=1 + cos x
(1 + cos x)2
=1
1 + cos x
(e) y =cos2 xsin2 x
Solution:
u = cos2 x v = sin2 x
⇒dudx
= −2 cos x sin x ⇒dvdx
= 2 sin x cos x
⇒dydx
=sin2 x × −2 cos x sin x − cos2 x × 2 sin x cos x
sin4 x
=−2 cos x sin x
(sin2 x + cos2 x
)sin4 x
=−2 cos x sin x
sin4 x
=−2 cos x
sin3 x
Solutions - Differentiation with Trig Page 8 of 26
(f) y =sin 2x
2 + cos 2x
Solution:
u = sin 2x v = 2 + cos 2x
⇒dudx
= 2 cos 2x ⇒dvdx
= −2 sin 2x
⇒dydx
=(2 + cos 2x) × 2 cos 2x − sin 2x × −2 sin 2x
(2 + cos 2x)2
=4 cos 2x + 2 cos2 2x + 2 sin2 2x
(2 + cos 2x)2
=4 cos 2x + 2(2 + cos 2x)2
(g) y =sin 3x
ex
Solution:
u = sin 3x v = ex
⇒dudx
= 3 cos 3x ⇒dvdx
= ex
⇒dydx
=ex × 3 cos 3x − sin 3x × ex
e2x
=3ex cos 3x − ex sin 3x
e2x
=3 cos 3x − sin 3x
ex
Solutions - Differentiation with Trig Page 9 of 26
4 Find Higher Derivatives
4. Find the second derivative of each of the following:
(a) f (x) = sin x
Solution:
f ′ (x) = cos xf ′′ (x) = − sin x
(b) f (x) = cos x
Solution:
f ′ (x) = − sin xf ′′ (x) = − cos x
(c) f (x) = tan x
Solution:
f ′ (x) =1
cos2 x= (cos x)−2
f = u−2 u = cos x
f ′ = −2u−3 u′ = − sin x
f ′′ (x) = −2u−3 × − sin x
=2 sin xcos3 x
Solutions - Differentiation with Trig Page 10 of 26
(d) f (x) = sin (2x)
Solution:
f ′ (x) = 2 cos (2x)f ′′ (x) = −4 sin (2x)
(e) f (x) = cos (3x)
Solution:
f ′ (x) = −3 sin (3x)f ′′ (x) = −9 cos (3x)
(f) f (x) = sin2 x
Solution:
f ′ (x) = 2 sin x cos x
f ′′ (x) = 2 cos2 x − 2 sin x(see product rule)
(g) f (x) = cos3 x
Solution:
f ′ (x) = −3 cos2 x sin x
u = −3 cos2 x v = sin x
⇒dudx
= 6 cos x sin x ⇒dvdx
= cos x
f ′′ (x) = −3 cos2 x × cos x + sin x × 6 cos x sin x
= −3 cos3 x + 6 cos x sin2 x
Solutions - Differentiation with Trig Page 11 of 26
(h) f (x) = tan2 (2x)
Solution:
f = u2 u = tan v v = 2x
⇒ f ′ = 2u ⇒ u′ =1
cos2 v⇒ v′ = 2
f ′ (x) = 2 tan v ×1
cos2 v× 2
=4 tan (2x)cos2 (2x)
=4 sin (2x)cos3 (2x)
(not necessary, but sin/cos are easier than tan for next step)
u = 4 sin (2x) v = cos3 (2x)
⇒dudx
= 8 cos (2x) ⇒dvdx
= −6 cos2 (2x) sin (2x)
f ′′ (x) =cos3 (2x) × 8 cos (2x) − 4 sin (2x) × −6 cos2 (2x) sin (2x)
cos6 (2x)
=8 cos4 (2x) + 24 sin2 (2x) cos2 (2x)
cos6 (2x)
=8 cos2 (2x) + 24 sin2 (2x)
cos4 (2x)
=1 + 16 sin2 (2x)
cos4 (2x)(not really necessary but looks clean)
5 Solve Problems about Gradients
5. QB T6P1 Q1 Let g (x) = 2x sin x
(a) Find g′ (x).
Solution:
g′ (x) = 2 sin x + 2x cos x
Solutions - Differentiation with Trig Page 12 of 26
(b) Find the gradient of the graph of g at x = π.
Solution:
g′ (π) = 2 sin π + 2π cos π= 0 − 2π = −2π
6. QB T6P1 Q9 Let h (x).
Find h′ (0)
Solution:
h′ (x) =6 cos x + 6x sin x
cos2 x
⇒ h′ (0) =6 cos 0 + 6 (0) sin 0
cos2 0
=612
= 6
7. QB T6P4 Q20 Consider f (x) = x2 sin x
(a) Find f ′ (x).
Solution:
f ′ (x) = 2x sin x + x2 cos x
(b) Find the gradient of the curve of f at x = π2 .
Solution:
f ′(π
2
)= 2
(π
2
)sin
π
2+
(π
2
)2cos
π
2= π + 0= π
Solutions - Differentiation with Trig Page 13 of 26
8. QB T6P2 Q4 Let f (x) = e2x cos x, −1 ≤ x ≤ 2.
(a) Show that f ′ (x) = e2x (2 cos x − sin x).
Solution:
u = e2x v = cos x
⇒dudx
= 2e2x ⇒dvdx
= − sin x
⇒ f ′ (x) = e2x × − sin x + cos x × 2e2x
= −e2x sin x + 2e2x cos x
= e2x (2 cos x − sin x)
(b) Find the equation of L.
Solution:
mT = f ′ (0) = e2×0 (2 cos 0 − sin 0)
= e0 (2 − 0)= 2
⇒ mL = −12
f (0) = e2×0 cos 0= 1
y − y1 = m (x − x1)
⇒ y − 1 = −12
(x − 0)
⇒ y = −x2
+ 1
Solutions - Differentiation with Trig Page 14 of 26
(c) Find the x-coordinate of P.
Solution:
e2x cos x = −x2
+ 1
9. 11M.2.sl.TZ1.8(a) Show that a = 4.
Solution: Amplitude is half the diameter⇒ a = 4.
(b) Show that b = π15 .
Solution:
T = 30
⇒ b =2π30
=π
15
Solutions - Differentiation with Trig Page 15 of 26
(c) Find these values of t.
Solution:
h′ (t) = −0.5 =4π15
cosπ
15t
⇒ −0.5 ×154π
= cosπ
15t
⇒ −0.5968310... = cosπ
15t
⇒π
15t = 2.210342088 or ⇒
π
15= 4.072843219
⇒ t = 10.55s ⇒ t = 19.45
(d) Determine whether the bucket is underwater at the second value of t.
Solution:
h (19.45) = 4 sin(π
15(19.45)
)+ 2
= −1.211 < 0⇒ under water
6 Solve Problems about Extrema
10. Find the x-coordinate of all extrema of each of the following functions for 0 ≤ x ≤ 2π.
Really need to be on top of radians and unit circle to tackle these questions effectively. Inparticular, note how having a function within a trig function requires a lot of consideration ofextra cycles and reverse cycles of the unit circle to obtain all valid solutions. For sin (bx + c)(or cos), there should be ≈ 2b solutions. Solutions below will skip some aspects of differenti-ation as a time-saver.
(a) cos x
Solution:
f ′ (x) = − sin x = 0⇒ x = 0, π, 2π
Solutions - Differentiation with Trig Page 16 of 26
(b) sin (2x)
Solution:
f ′ (x) = 2 cos (2x) = 0
⇒ 2x =π
2,
3π2,
5π2,
7π2
⇒ x =π
4,
3π4,
5π4,
7π4
(c) cos (3x)
Solution:
f ′ (x) = −3 sin (3x) = 0⇒ 3x = 0, π, 2π, 3π, 4π, 5π, 6π
⇒ x = 0,π
3,
2π3, π,
4π3,
5π3, 2π
(d) sin(2x + π
2
)Solution:
f ′ (x) = 2 cos(2x +
π
2
)= 0
⇒ 2x +π
2= −
π
2,
pi2,
3π2,
5π2,
7π2,
9π2
⇒ 2x = −π, 0, π, 2π, 3π, 4π
⇒ x =���−π
2, 0,
π
2, π,
3π2, 2π
(e) cos(
12 x − π
)Solution:
f ′ (x) = −12
sin(12
x − pi)
= 0
⇒12
x − π = −π, 0, π, 2π, 3π
⇒12
x = ���−2π, 0, 2π,��4π,��6π
Solutions - Differentiation with Trig Page 17 of 26
(f) 2 sin (4x + 2π)
Solution:
f ′ (x) = 8 cos (4x + 2π) = 0
⇒ 4x + 2π = −π
2,π
2,
3π2,
5π2,
7π2,
9π2,
11π2,
13π2,
15π2,
17π2,
19π2,
21π2
⇒ 4x = −5π2,−
3π2,−π
2,π
2,
3π2,
5π2,
7π2,
9π2,
11π2,
13π2,
15π2,
17π2
⇒ x =���
−5π8,���
−3π8,���−π
8,π
8,
3π8,
5π8,
7π8,
9π8,
11π8,
13π8,
15π8,���17π
8
11. QB T6P3 Q21(a) Use the graph to write down the value of a, c, and d.
Solution:
a =12
(12 − −4) = 8c = 4 −14
wavelength = 2d = 12 − a = 4
(b) Show that b = π4 .
Solution:
b =2πT
=2π8
=π
4
(c) Find f ′ (x).
Solution:
f (x) = 8 sin(π
4(x − 2)
)+ 4 = 8 sin
(π
4x −
π
2
)+ 4⇒ f ′ (x) = 2π cos
(π
4(x − 2)
)
Solutions - Differentiation with Trig Page 18 of 26
(d) Find the x-coordinate of R.
Solution:
2π cos(π
4(x − 2)
)= −2π
⇒ cos(π
4(x − 2)
)= −1
⇒π
4(x − 2) = −π, π, 3π...
⇒π
4x −
π
2= −π, π, 3π...
⇒π
4x = −
π
2,
3π2,
7π2...
⇒ x = ��−2, 6,��14
7 Solve Problems about Points of Inflexion
12. Find the x-coordinate of all points of inflexion of each of the following functions for 0 ≤ x ≤2π.
(a) cos x
Solution:
f ′′ (x) = − cos x
⇒ x =π
2,
3π2
(b) tan x
Solution:
f ′′ (x) =2 sin xcos3x
⇒ x = 0, π, 2π
(c) sin (2x)
Solution:
f ′′ (x) = −4 sin (2x)
⇒ x = 0,π
2, π,
3π2, 2π
Solutions - Differentiation with Trig Page 19 of 26
(d) cos (3x)
Solution:
f ′′ (x) = −9 cos (3x)
⇒ x =π
6,π
2,
5π6,
7π6,
3π2,
11π6
(e) tan (2x)
Solution:
f ′′ (x) =8 sin (2x)cos3 (2x)
⇒ x = 0,π
2, π,
3π2, 2π
(f) sin(2x + π
2
)Solution:
f ′′ (x) = −4 sin(2x +
π
2
)⇒ x =
π
4,
3π4,
5π4,
7π4
(g) cos(
12 x − π
)Solution:
f ′′ (x) = −14
cos(12
x − π)
⇒ x = π
(h) 2 sin (4x + 2π)
Solution:
f ′′ (x) = −32 sin (4x + 2π)
⇒ x = 0,π
4,π
2,
3π4, π,
5π4,
3π2,
7π4, 2π
Solutions - Differentiation with Trig Page 20 of 26
13. 10M.1.sl.TZ1.9(a) Use the quotient rule to show that f ′ (x) = − 1
sin2 x.
Solution:
u = cos x v = sin x
⇒dudx
= − sin x ⇒dvdx
= cos x
⇒ f ′ (x) =(sin x) (−sinx) − (cos x) (cos x)
sin2 x
=− sin2 x − cos2 x
sin2 x
=−1
(sin2 x + cos2 x
)sin2 x
= −1
sin2 x
(b) Find f ′′ (x).
Solution:
f ′ (x) = sin−2 x
⇒ f ′′ (x) =2 cos xsin3 x
(c) Find the values of p and q.
Solution:
f ′(π
2
)= −
1sin2 π
2
= −1
f ′′(π
2
)=
2 cos π2
sin3 π2
= 0
(d) Use the information in the table to explain why there is a point of inflexion on the graphof f where x = π
2 .
Solution: f ′′ (x) has a root at π2 and f changes from concave up to concave down at
π2 , indicating that it is a point of inflexion.
Solutions - Differentiation with Trig Page 21 of 26
8 Solve Optimization Problems
14. 11M.2.sl.TZ2.10(a) Show that the area of the window is given by y = 4 sin θ + 2 sin 2θ.
Solution:
Area =a + b
2h a = 2
y =2 + 2 + 4 cos θ
2× 2 sin θ h = 2 sin θ
=4 + 4 cos θ
2× 2 sin θ b = 2 + 4 cos θ
= (2 + 2 cos θ) (2 sin θ)= 4 sin θ + 2 (2 cos θ sin θ)= 4 sin θ + 2 sin 2θ
(b) Zoe wants a window to have an area of 5 m2. Find the two possible values of θ.
Solution:
4 sin θ + 2 sin 2θ = 5
From graph, θ ≈ 0.85, 1.25
Solutions - Differentiation with Trig Page 22 of 26
(c) Find all possible values for A.
Solution: From graph, it looks like approximately 4 < A < 5.2.
y(π
2
)= 4 sin
(π
2
)+ 2 sin
(2π
2
)= 4 (1) + 2 (0) = 4
, so minimum value is 4.For maximum value, need derivative = 0:
y′ (θ) = 4 cos θ + 4 cos 2θ = 0⇒ cos θ + cos 2θ = 0
⇒ cos θ + 2 cos2 θ − 1 = 0 cos 2θ = 2 cos2 θ − 1
⇒ 2 cos2 θ + cos θ − 1 = 0
⇒ 2u2 + u − 1 = 0 u = cos θ⇒ (2u − 1) (u + 1) = 0
⇒ u =12
⇒ u = −1
⇒ cos θ =12
⇒ cos θ = −1
⇒ θ =π
3⇒ θ =�π
y(π
3
)= 4 sin
(π
3
)+ 2 sin
(2π
3
)≈ 5.20
, so possible values for A are 4 ≤ A < 5.20
15. 08M.1.sl.TZ2.10(a) Find the area of the triangle OPB in terms of θ.
Solution:
A =12
ab sin θ
=12
(2) (2) sin θ
= 2 sin θ
Solutions - Differentiation with Trig Page 23 of 26
(b) Explain why the area of the triangle OPA is the same as the area of triangle OPB.
Solution: They have the same base length (2) and the same height (as their uppervertex is in the same place).
(c) Show that S = 2 (π − 2 sin θ).
Solution:
S =12πr2 − 2 (2 sin θ)
=12π22 − 4 sin θ
= 2π − 4 sin θ= 2 (π − 2 sin θ)
(d) Find the value of θ when S is a local minimum, justifying that it is a local minimum.
Solution:
S ′ (θ) = −4 cos θ = 0⇒ cos θ = 0
⇒ θ =π
2
S(π
2
)= 2
(π − 2 sin
(π
2
))= 2 (π − 2)≈ 2.28
(e) Find a value of θ for which S has its greatest value.
Solution:
θ = 0, π
Solutions - Differentiation with Trig Page 24 of 26
9 Solve Kinematic Problems
16. 12N.2.sl.TZ0.7(a) Sketch the graph of s.
Solution:
(b) Find the maximum velocity of the particle.
Solution:
⇒ v (t) = −2t cos t + 2 cos t⇒ v′ (t) = −2t cos t − 2 sin t − 2 sin t = 0
⇒ −2t cos t − 4 sin t = 0
⇒ t − 2sin tcos t
= 0
⇒ t − 2 tan t = 0
This is not solvable with our level of mathematics, so we try a graph instead:
Solutions - Differentiation with Trig Page 25 of 26
Thus vmax ≈ 10.2ms−1
17. 12M.1.sl.TZ1.10(a) Find s′ (t)
Solution:
s′ (t) = 1 − 2 cos (2t)
(b) Find the other value.
Solution:
1 − 2 cos (2t) = 0
⇒ cos (2t) =12
⇒ 2t =π
3,
5π3
⇒ t =���π
6,
5π6
(c) Show that s′ (t) > 0 between these two values of t.
Solution:
1 − 2 cos (2π)= 1 − 2 (−2)= 5 > 0
Solutions - Differentiation with Trig Page 26 of 26