MCV 4U1 Grade 12 Calculus and Vectors
Derivatives of Trigonometric and Exponential Functions
Review
1. Differentiate each of the following:
a) y = 6 – ex
b) y = 2x + 3e –x
d) y = 𝑒 −3𝑥
c) y = e 2x + 3
x
e) y = x e
f) s =
Answer: a) – ex
𝑑𝑦
𝑑𝑥
𝑒𝑡 − 1
𝑒𝑡 + 1
b) 2 + 3ex
d) (– 6x + 5) 𝑒
2. Determine
2 + 5𝑥
−2𝑥 2 + 5𝑥
c) 2e2x + 3
x
e) (x + 1) e
for each of the following:
2
a) y = 10x
b) y = 43𝑥
c) y = (5x)5x
d) y = (x4)2x
2𝑒 𝑡
f) (𝑒 𝑡
+ 1)2
e) y =
4𝑥
f) y =
4𝑥
Answer:
a) 10x ln 10
c) 5x + 1 (x ln 5 + 1)
4 − 4𝑥 ln 4
e)
𝑥
4
1
2 − cos 𝑥
e) y = sin 2x e3x
𝑥
2
b) (6 ln 4) (x43𝑥 )
d) x32x (x ln 2 + 4)
1
ln 5
f) 5√𝑥 (− 2 +
)
3. Differentiate each of the following:
a) y = 3sin2x – 4cos2x
c) y =
5 √𝑥
𝑥
2𝑥√𝑥
b) y = tan 3x
d) y = x tan 2x
f) y = cos2 2x
Answer:
a) 6 cos 2x + 8 sin 2x
sin 𝑥
c) – (2−cos 2
b) 3 sec2 3x
d) 2x sec2 2x + tan 2x
𝑥)
f) –2 sin 4x
3x
e) e (3sin 2x + 2 cos 2x)
𝑒𝑥
4. Given the function f (x) = , solve the equation f' (x) = 0. Discuss the
𝑥
significance of the solution you found.
Solution
(𝑒 𝑥 )′ (𝑥) − (𝑒 𝑥 )(𝑥)′
f' (x) =
𝑥2
f' (x) = 0 ⇒ x = 1.
=
𝑒 𝑥 (𝑥 − 1)
𝑥2
The function has a horizontal tangent at (1, e). It is a local minimum point.
f' (x) –
+
↘
1
↗
5. If f (x) = xe –2x, find f ' (0.5). Explain what this number represents.
Solution
f' (x) = e –2x – 2xe –2x = (1 – 2x) e –2x
f ' (0.5) = 0 ∴ The slope of the tangent to at the point with x-coordinate 0.5
is 0.
6. Determine the second derivative of each of the following:
a) y = xex – ex
b) y = xe10x
Answer: a) ex (x + 1)
b) 20 e10x (5x + 1)
𝑒 2𝑥 − 1
7. If y = 2𝑥
𝑒 +
Solution
1
prove that
𝑑𝑦
𝑑𝑥
= 1 – y2.
8. Determine the equation of the tangent to the curve defined by y = x – e – x
that is parallel to the line represented by 3x – y – 9 = 0.
Solution
Determine the slope of the tangent:
3x – y – 9 = 0 ⇒ y = 3x – 9 ⇒ m = 3
Determine the tangency point:
y' = 1 + e – x
y' = m ⇒ 1 + e – x = 3 ⇒ e – x = 2 ⇒ – x = ln 2 ⇒ x = – ln 2
y = x – e – x = – ln 2 – 2
The tangency point is: (– ln 2, – ln 2 – 2)
Determine the equation of the tangent:
y = m (x – x1) + y1 ⇒ y = 3(x + ln 2) – ln 2 – 2
y = 3x + 2 ln 2 – 2
9. Determine the equation of the tangent to the curve y = x sin x at the point
𝜋
where x = .
2
Solution
𝜋
𝜋
𝜋
𝜋
𝜋 𝜋
y = y( ) = ( ) sin( ) = ∴ the tangency point is: ( , )
2
2
2
2
2 2
y' = sin x + x cos x
𝜋
𝜋
𝜋
𝜋
m = y' ( ) = sin ( ) + ( ) cos ( ) = 1
2
2
2
2
𝜋
𝜋
2
2
y = m (x – x1) + y1 ⇒ y = 1(𝑥 – ) + ⇒ y = x.
10. An object moves along a line so that, at time t, in seconds, its position is
sin 𝑡
s=
,
3 + cos 2𝑡
where s is the displacement in metres. Calculate the object's velocity at t =
𝜋
.
4
Solution
s' =
=
(sin 𝑡)′ (3 + cos 2𝑡)− (sin 𝑡)(3 + cos 2𝑡)′
(3 + cos 2𝑡)2
(cos 𝑡) (3 + cos 2𝑡) − (sin 𝑡)(−2 sin 2𝑡)
(3 + cos 2𝑡)2
𝜋
𝜋
𝜋
cos ( 4 )(3 + cos ( 2 )) − sin( 4 )(−2 sin ( 2 ))
4
(3 + cos( 2 ))
v = s' ( ) =
=
𝜋
𝜋
2
√2
(3 + 0) − √ (−2)
2
2
(3 + 0)2
𝜋
=
5√2
18
2
≈ 0.3928 m/s
The object's velocity is about 0.3928 m/s.
11. The number of bacteria in a culture, N, at time t days is given by
−
𝑡
20
N (t) = 2000(30 + 𝑡𝑒 )
a) When is the rate of change of the number of bacteria equal to zero?
Solution
N' (t) = 2000(𝑒
𝑡
−20
N' (t) = 0 ⇒ 1 –
𝑡
20
−
𝑡
20
𝑒
𝑡
−20
) = 2000𝑒
𝑡
−20
(1 −
𝑡
20
)
= 0 ⇒ t = 20 days.
b) If the bacterial culture is placed in a colony of mice, the number of mice
that become infected, M, is related to the number of bacteria present by the
3
equation M (t) = √𝑁 + 1000. After 10 days, how many mice are infected
per day?
Solution
1
M (t) = (𝑁(𝑡) + 1000)3
1
2
−3
M' (t) = (𝑁 (𝑡) + 1000)
3
∙ N' (t)
1
−
2
3
M' (10) = (𝑁(10) + 1000) ∙ N' (10)
3
N (10) = 2000(30 + 10𝑒 −0.5 ) ≈ 72130.6132
𝑡
N' (10) = 2000𝑒 −20 (1 −
1
𝑡
) ≈ 606.53
20
2
−
M' (10) = (73130.6132) 3 ∙ (606.53) ≈ 0.1156
3
After 10 days, about 0.1156 mice are infected per day.
12. The concentrations of two medicines in the bloodstream t hours after
injection are c1 (t) = te – t and c2 (t) = t2 e – t.
a) Which medicine has the larger maximum concentration?
Solution
c1' (t) = te – t = e – t – te – t = e – t (1 – t)
c1' (t) = 0 ⇒ e – t (1 – t) = 0 ⇒ t = 1
t
0≤t<1 1 t >1
+
0
–
c1' (t)
–1
c1 (t)
e
↗
↘
max
1
∴ max c1 (t) = c1 (1) =
𝑡 ≥0
𝑒
c2' (t) = 2t e – t e = t e – t (2 – t)
c2' (t) = 0 ⇒ t = 0 or t = 2
t
c2' (t)
c2(t)
–t
0
0
0
2
–t
0<t<2
+
↗
2
t>2
0
–
4e – 1
↘
max
4
∴ max c2(t) = c2 (2) = ⇒ max c2(t) > max c1 (t)
𝑒
𝑡 ≥0
𝑡 ≥0
𝑡 ≥0
b) Within the first half hour, which medicine has the larger maximum
concentration?
Solution
max c1 (t) = c1 (0.5) = 0.5e – 0.5
0≤𝑡≤0.5
max c2 (t) = c2 (0.5) = 0.25e – 0.5
0≤𝑡≤0.5
∴ max c1 (t) > max c2 (t).
0≤𝑡≤0.5
0≤𝑡≤0.5
13. Differentiate.
a) y = (2 + 3e –x)3
b) y = xe
c) y = 𝑒 𝑒
𝑥
d) y = (1 – e5x)5
Answer:
a) y = – 9 e –x (2 + 3e –x)2
b) e xe - 1
𝑥
c) 𝑒 𝑥 + 𝑒
d) – 25 e5x (1 – e5x)4
14. Differentiate.
a) y = 5x
b) y = 0.4x
c) y = 52x
d) y = 5 (2x)
e) y = 4 (ex)
f) y = – 2(103x)
Answers
15. Determine y'.
a) y = sin 2x
b) y = x2 sin x
c) y = sin (π/2 – x)
d) y = cos x ∙ sin x
e) y = cos2 x
f) y = cos x sin2 x
Answers
𝜋
16. Determine the equation of the tangent to the curve y = cos x at ( , 0).
2
Solution
y' = – sin x
y' (π/2) = – sin (π/2) = –1
The equation of the tangent is:
y = m(x – x1) + y1
y = – (x – π/2) + 0
x + y – π/2 = 0
17. An object is suspended from the end of a spring. Its displacement from
the equilibrium position is s = 8 sin (10πt) at time t. Calculate the velocity
𝑑2 𝑠
and acceleration of the object at any time t, and show that 2 + 100π2s = 0.
𝑑𝑡
Solution
𝑑𝑠
v (t) = = 8 cos (10πt) ∙ (10πt)' = 80π cos (10πt)
𝑑𝑡
𝑑𝑠
𝑑2 𝑠
a (t) = = 2 = –80π sin (10πt) ∙ (10πt)'
𝑑𝑡 𝑑𝑡
= –800 π2 sin (10πt) = –100 π2 [8 sin (10πt)] = –100 π2s
∴
𝑑2 𝑠
𝑑2 𝑠
𝑑𝑡
𝑑𝑡 2
2
2 = –100 π s ⇒
+ 100π2s = 0.
18. Determine f "(x).
a) f (x) = 4 sin2 (x – 2)
Solution
f ' (x) = 8 sin (x – 2) cos (x – 2) = 4 sin (2x – 2)
f '' (x) = 8 cos (2x – 2)
b) f (x) = 2 cos x sec2 x
Solution
f (x) = 2 cos x sec2 x = 2 (cos x ∙ sec x) sec x = 2 sec x
f ' (x) = 2 tan x ∙ sec x
f '' (x) = 2 [(tan x)' ∙ sec x + tan x ∙ (sec x)']
= 2 (sec2 x ∙ sec x + tan2 x ∙ sec x)
= 2 sec x ∙ (sec2 x + tan2 x)
19. The position of a particle is given by s = 5cos (2t + π/4) at time t. What
are the maximum values of the displacement, the velocity, and the
acceleration?
Solution
–1 ≤ cos (2t + π/4) ≤ 1 ⇒ –5 ≤ 5cos (2t + π/4) ≤ 5
∴ the maximum values of the displacement is 5.
𝑑𝑠
v (t) = = –10 sin (2t + π/4)
𝑑𝑡
–1 ≤ sin (2t + π/4) ≤ 1 ⇒ –10 ≤ –10 sin (2t + π/4) ≤ 10
∴ the maximum values of the velocity is 10.
𝑑𝑠
𝑑2 𝑠
a (t) = = 2 = –20 cos (2t + π/4)
𝑑𝑡 𝑑𝑡
–1 ≤ cos (2t + π/4) ≤ 1 ⇒ –20 ≤ –20 cos (2t + π/4) ≤ 20
∴ the maximum values of the acceleration is 20.
20. The hypotenuse of a right triangle is l2 cm in length. Calculate the
measures of the unknown angles in the triangle that will maximize its
perimeter.
Solution
P (𝛳) = 12 + x + y
= 12 + 12 cos 𝛳 + 12 sin 𝛳,
0 ≤ 𝛳 ≤ π/2
P' (𝛳) = –12 sin 𝛳 + 12 cos 𝛳
P' (𝛳) = 0 ⇒ –12 sin 𝛳 + 12 cos 𝛳 = 0
sin 𝛳 = cos 𝛳
tan 𝛳 = 1⇒ 𝛳 = π/4
P (0) = 12 + 12 cos 0 + 12 sin 0 = 24
P (π/2) = 12 + 12 cos (π/2) + 12 sin (π/2) = 24
P (π/4) = 12 + 12 cos (π/4) + 12 sin (π/4) = 12 + 12√2
∴ max P (𝛳) = P (π/4) = 12 + 12√2
0 ≤ ϴ ≤ π/2
The measure of the acute angle 𝛳 in the right triangle that maximizes its
perimeter is π/4.
Note: The problem could be solved without applying derivative:
P (𝛳) = 12 + 12 cos 𝛳 + 12 sin 𝛳
1
1
= 12 + 12√2 ( cos 𝛳 + sin 𝛳)
√2
√2
π
= 12 + 12√2
4 cos 𝛳 + cos /4 sin 𝛳)
= 12 + 12√2 sin (π/4 + 𝛳)
sin (π/4 + 𝛳) ≤ 1
P (𝛳) would be maximal when sin (π/4 + 𝛳) = 1
∴ π/4 + 𝛳 = π/2, 𝛳 = π/4
∴ max P (𝛳) = P (π/4) = 12 + 12√2
(sin π/
0 ≤ ϴ ≤ π/2
21. A fence is 1.5 m high and is 1 m from a wall. A ladder must start from
the ground, touch the top of the fence, and rest somewhere on the wall.
Calculate the minimum length of the ladder.
Solution
From △ABD determine AD:
𝐴𝐵
cot 𝛳 =
⇒ AB = 1.5 cot 𝛳
𝐵𝐷
AC = AB + BC = 1.5 cot 𝛳 + 1
From △ACE determine AE:
𝐴𝐶
𝐴𝐶
1.5 cot ϴ + 1
cos 𝛳 = ⇒ AE =
=
𝐴𝐸
cos 𝛳
=
cos 𝛳
1.5
1
sin 𝛳
cos 𝛳
+
∴ the length of the ladder equals:
1.5
1
L (𝛳) =
+
sin 𝛳 cos 𝛳
Determine the minimal value of L (𝛳).
L' (𝛳) = –1.5
cos 𝛳
sin2 𝛳
+
sin 𝛳
cos2 𝛳
=
–1.5 cos3 𝛳 + sin3 𝛳
sin2 𝛳 ∙ cos2 𝛳
L' (𝛳) = 0 ⇒ –1.5 cos3 𝛳 + sin3 𝛳 = 0
tan3 𝛳 = 1.5
𝟑
tan 𝛳 = √1.5 ≈ 1.1447
𝟑
𝛳 = tan –1 ( √1.5) ≈ 48.86o
L'
(45o)
L'
(50o)
L (𝛳) =
=
=
–1.5 cos3 45° + sin3 45°
sin2 45° ∙ cos2 45°
–1.5 cos3 50° + sin3 50°
sin2 50° ∙ cos2 50°
1.5
1
sin 48.86°
+
cos 48.86°
<0
>0
≈ 3.5 m
48.86o
𝛳 > 48.86o
0
+
3.5
↗
min
∴ the minimum length of the ladder is about 3.5 m.
0° < 𝛳 < 48.86o
𝛳
–
L' (𝛳)
L (𝛳)
↘
22. A thin rigid pole needs to be carried horizontally around a corner
joining two corridors, which are I m and 0.8 m wide. Calculate the length of
the longest pole that can be carried around this corner.
Solution
0.8
1
L (𝛳) = y + x =
+
sin 𝛳 cos 𝛳
Determine the minimal value of L (𝛳).
L' (𝛳) = – 0.8
cos 𝛳
sin2 𝛳
+
sin 𝛳
cos2 𝛳
=
–0.8 cos3 𝛳 + sin3 𝛳
sin2 𝛳 ∙ cos2 𝛳
L' (𝛳) = 0 ⇒ – 0.8 cos3 𝛳 + sin3 𝛳 = 0
tan3 𝛳 = 0.8
𝟑
tan 𝛳 = √0.8 ≈ 0.9283
𝟑
𝛳 = tan –1 ( √0.8) ≈ 42.87o
L'
(42o)
L'
(45o)
L (𝛳) =
=
=
–0.8 cos3 42° + sin3 42°
sin2 42° ∙ cos2 42°
–0.8 cos3 45° + sin3 45°
sin2 45° ∙ cos2 45°
0.8
1
sin 42.87°
+
cos 42.87°
<0
>0
≈ 2.5 m
42.87o
𝛳 > 42.87o
0
+
2.5
↗
min
∴ the minimum length of the pole is about 2.5 m.
∴ the longest pole that can be carried around this corner is about 2.5 m.
0° < 𝛳 < 42.87o
𝛳
–
L' (𝛳)
L (𝛳)
↘
23. When the rules of hockey were developed, Canada did not use the
metric system. Thus, the distance between the goal posts was designated to
be six feet (slightly less than 2 m). If Sidney Crosby is on the goal line,
three feet outside one of the goal posts, how far should he go out
(perpendicular to the goal line) to maximize the angle in which he can shoot
at the goal?
Hint: Determine the values of x that maximize 𝜃 in the following diagram.
Solution
By the Pythagorean Theorem, we get:
a2 = x2 + 9 ⇒ a = √𝑥 2 + 9
b2 = x2 + 81 ⇒ b = √𝑥 2 + 81
By the Cosine Law, we get:
36 = a2 + b2 – 2ab cos 𝛳 ⇒ cos 𝛳 =
𝑥 2 + 9 + 𝑥 2 + 81 − 36
𝑎2 + 𝑏2 − 36
2𝑎𝑏
𝑥 2 + 27
cos 𝛳 =
⇒ cos 𝛳 = 4
√𝑥 + 90𝑥 2 + 729
2√𝑥 2 + 9 ∙ √𝑥 2 + 81
To maximize 𝜃, we should minimize cos 𝛳.
Determine the values of x that maximize
4
1
−2
2
f (x) = (x + 27) (𝑥 + 90𝑥 + 729 ) , x > 0
2
1
f '(x) = (x2 + 27)' (𝑥 4 + 90𝑥 2 + 729 )−2 +
4
1
−2
2
′
(x + 27) [(𝑥 + 90𝑥 + 729 ) ]
2
1
= 2x (𝑥 4 + 90𝑥 2 + 729)−2
1
2
4
2
– (𝑥 + 27)(𝑥 + 90𝑥 + 729 )
2
3
−2
(4x3 + 180x)
1
= 2x (𝑥 4 + 90𝑥 2 + 729)−2
2
4
2
– (𝑥 + 27)(𝑥 + 90𝑥 + 729 )
3
−2
(2x3 + 90x)
4
2
= 2x (𝑥 + 90𝑥 + 729 )
= 2x (𝑥 4 + 90𝑥 2 + 729 )
4
3
2
3
−2
−
[x4 + 90x2 + 729 – (x2 + 27)(x2 + 45)]
(18x2 – 486)
3
−2
2
= 36x (𝑥 + 90𝑥 + 729 ) (x2 – 27)
f '(x) = 0, x > 0 ⇒ x2 = 27 ⇒ x = 3√3 ≈ 5.2 ft
x 0 < x < 3√3
f '(x)
–
f (x)
↘
3√3
0
min
x > 3√3
+
↗
∴ 𝐦𝐢𝐧 f (x) = f (3√3) ⇒𝐦ax 𝛳 (x) = 𝛳 (3√3)
𝑥>0
𝑥>0
To maximize the angle in which he can shoot at the goal, Sidney Crosby
should he go out (perpendicular to the goal line) about 5.2 ft.