AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule...
Transcript of AS Level -Level Differentiation Pt. 3: General Differentiation...Differentiation Pt. 5: Product Rule...
Differentiation
Pt. 3: General Differentiation
1. Find π2π¦
ππ₯2 for y = 4x3 + ex (2)
2. Find ππ¦
ππ₯ for y =
3
π₯+ 3 ln π₯ (2)
3. Find any values of x for which ππ¦
ππ₯ = 0 when y = 5ln x β 8x (2)
4. Find the co-ordinates and the nature of any stationary points on the curve y = 7 + 2x β 4 lnx (5)
5. Find an equation for the normal to the cube y = 10 + ln x at the x-cordinate 3. (4)
6. Find the differential of 4sin x + 2 cos x (2)
7. Differentiate cot(2x β 3) (1)
8. Find an equation for the tangent for the curve y = cosec x β 2sin x at the point x = π
6 in the interval 0 β€ x β€ 2π (5)
9. Find the differential of y = sec 5x (1)
10. Differentiate with respect to x, 3x
11. Differentiate with respect to x, 51-x
12. Differentiate with respect to x, 2π₯3
A-Level
Pt. 3: General Differentiation
Pt. 4: The Chain Rule
Pt. 5: The Product Rule
Pt. 6: The Quotient Rule
Pt. 7: Parametric Differentiation
Pt. 8: Implicit Differentiation
Pt. 9: Rates of Change
AS Level
Pt. 1: First Principles
Pt. 2: Differentiation
Pt. 9: Rates of Change
Pt. 1: First Principles
Pt. 2: Differentiation
Mark Scheme
1.
ππ¦
ππ₯= 12x2 + ex M1
π2π¦
ππ₯2 = 24x + ex M1
2.
y = 3
π₯+ 3 ln π₯ = 3x-1 + 3 ln x M1
ππ¦
ππ₯ = -3x-2 +
3
π₯ M1
3.
ππ¦
ππ₯ =
5
π₯ - 8 M1
5
π₯ β 8 = 0
5
π₯ = 8
x = 5
8
M1
3.
ππ¦
ππ₯ = 2 -
4
π₯ M1
Stationary point: 2 - 4
π₯ = 0
x = 2 M1
π2π¦
ππ₯2 = 4x-2 M1
x = 2, π2π¦
ππ₯2 = 1
π2π¦
ππ₯2 > 0, therefore minimum point M1
x = 2, y = 11 β 4ln 2 M1
Therefore (2, 11 β 4ln 2) is a minimum point
5.
x = 3, y = 10 + ln 3 M1
ππ¦
ππ₯=
1
π₯ M1
When x = 3, ππ¦
ππ₯ =
1
3
Gradient of normal = -3 M1
Therefore equation of line is:
y β (10 + ln 3) = -3(x β 3)
y = 19 + ln 3 β 3x
M1
6.
ππ¦
ππ₯= 4cos x β 2 sin x M1
7.
ππ¦
ππ₯= -2coscec2(2x β 3) M1
8.
πβππ π₯ =π
6, π¦ = 1 M1
ππ¦
ππ₯ = - cosec x cot x β 2 cos x M1
ππ¦
ππ₯ (
π
6) = β cosec (
π
6) cot (
π
6) β 2 cos (
π
6)
ππ¦
ππ₯ = 3β3
M1
y β 1 = 3β3 (x - π
6) M1
6β3π₯ + 2y β 2 - β3π = 0 M1
9.
ππ¦
ππ₯ = 5 sec 5x tan 5x M1
10.
π
ππ₯(3x) = 3x ln 3 M1
11.
π
ππ₯(51 β x) = 51 β x ln 5 Γ -1 M1
π
ππ₯(51 β x) = -(51 β x)ln 5
12.
π
ππ₯(2π₯3
) = 2π₯3ln 2 Γ 3x2 M1
π
ππ₯(2π₯3
) = 3x2(2π₯3) ln 2
Differentiation
Pt. 4: Chain Rule
1. Differentiate with respect to x, (x + 3)5 (1)
2. Find π2π¦
ππ₯2 when y = 4 ln (1 + 2x) (2)
3. Find the value of fβ(x) when f(x) = 4β3π₯ + 15 (2)
4. Find the coordinates of any stationary points for the curve y = 8x β e2x (4)
5. Find an equation for the tangent to the curve y = 2 + ln (1 + 4x) given the x-coorinate is 0. (4)
6. Find an equation for the normal to the curve y = 1
2βln π₯ when x = 1. (5)
7. Find the coordinates of any stationary points on the curve y = 2 ln(x β x2). (5)
8. A curve has the equation y = β8 β π2π₯
The point P on the curve has y-coordinate 2.
a. Find the x-coordinate of P. (2)
b. Show that the tangent to the curve at P has equation 2x + y = 2 + ln 4. (2)
9. A quantity N is increasing such that at time t seconds, N = aekt.
Given that at time t = 0, N = 20 and that at time t = 8, N = 60, find
a. The values of the constants a and k (3)
b. The value of N when t = 12 (2)
c. The rate at which N is increasing when t = 12. (3)
10. f(x) = (5 β 2x2)3
a. Find fβ(x) (2)
b. Find the coordinates of the stationary points of the curve y = f(x). (4)
c. Find the equation for the tangent to the curve y = f(x) at the point with x-coordinate 3
2 giving your answer in the
form ax + by + c = 0, where a, b and c are integers. (3)
11. The diagram shows the curve y = f(x) where f(x) = 3 ln 5x β 2x, x > 0.
a. Find f β²(x). (1)
b. Find the x-coordinate of the point on the curve at which the gradient of the normal to the curve is -1
4 . (2)
c. Find the coordinates of the maximum turning point of the curve. (3)
d. Write down the set of values of x for which f(x) is a decreasing function. (1)
12. f(x) β‘ x2 β 7x + 4 ln (π₯
2), x > 0.
a. Solve the equation f β²(x) = 0, giving your answers correct to 2 decimal places. (3)
b. Find an equation for the tangent to the curve y = f(x) at the point on the curve where x = 2. (2)
A-Level
Pt. 3: General Differentiation
Pt. 4: The Chain Rule
Pt. 5: The Product Rule
Pt. 6: The Quotient Rule
Pt. 7: Parametric Differentiation
Pt. 8: Implicit Differentiation
Pt. 9: Rates of Change
AS Level
Pt. 1: First Principles
Pt. 2: Differentiation
Pt. 9: Rates of Change
Pt. 1: First Principles
Pt. 2: Differentiation
Mark Scheme
1.
5(x + 3)4 M1
2. ππ¦
ππ₯=
4
1+2π₯ Γ 2 =
8
1+2π₯ = 8(1 + 2x)-1 M1
π2π¦
ππ₯2 = -8(1 + 2x)-2 x 2
π2π¦
ππ₯2 = β16
(1+2π₯)2 M1
3.
f(x) = 4(3x + 15)0.5 M1
fβ(x) = 2(3x + 15)-0.5 x 3
fβ(x) = 6(3x + 15)-0.5 M1
4. ππ¦
ππ₯ = 8 β 2e2x M1
8 β 2e2x = 0
2e2x = 8
e2x = 4
M1
2x = ln 4
x = 1
2ln 4 = ln 2
M1
When x = ln 2
y = 8ln 2 β 4 M1
5.
When x = 0
y = 2 + ln (1 + 0) = 2 M1
ππ¦
ππ₯=
1
1+4π₯ Γ 4 =
4
1+4π₯ M1
When x = 0, ππ¦
ππ₯ = 4 M1
y = 4x + 2 M1
6.
When x = 1,
y = 1
2βln 1 =
1
2
M1
ππ¦
ππ₯ = -(2 β ln x)-2 Γ -(
1
π₯) =
1
π₯(2βπππ₯)2 M1
When x = 1, ππ¦
ππ₯ =
1
1(2βln 1)2 = -4 M1
y β 1
2= β4(π₯ β 1) M1
y = 9
2β 4π₯ M1
7. ππ¦
ππ₯ =
2
π₯β π₯2 Γ (1 β 2π₯) = 2(1β2π₯)
π₯β π₯2 M1
At stationary point ππ¦
ππ₯ = 0
2(1β2π₯)
π₯β π₯2 = 0
2(1 β 2x) = 0
M1
x = 1
2 M1
When x = 1
2
y = 2 ln(1
2 β
1
22).
y = -4 ln 2
M1
Stationary point: (1
2, -4 ln 2) M1
8a.
β8 β π2π₯ = 2
8 β e2x = 4 M1
x = 1
2ln 4 = ln 2 M1
8b. ππ¦
ππ₯=
1
2(8 β e2x)-0.5 x (-2e2x) M1
ππ¦
ππ₯ =
βπ2π₯
β8β π2π₯ M1
When x = ln 2 ππ¦
ππ₯ =
βπ2ln 2
β8β π2ln 2 = -2
M1
y β 2 = -2(x β ln 2)
2x + y = 2 + 2 ln 2 M1
2x + y = 2 + ln 22
2x + y = 2 + ln 4 M1
9a.
t = 0, N = 20
Therefore a = 20 M1
t = 8, N = 60
60 = 20e8k M1
k = 1
8ln 3 = 0.137 M1
9b.
N = 20e0.1373t M1
t = 12, N = 104 M1
9c. ππ
ππ‘= 20 Γ 0.1373e0.1373t = 2.747e0.1373t M1
t = 12, ππ
ππ‘ = 14.3 M1
N is increasing at 14.3 per second (3 s.f) M1
10a. ππ¦
ππ₯ = 3(5 β 2x2)2 Γ (β4π₯) M1
ππ¦
ππ₯ = -12x(5 β 2x2)2 M1
10b.
At stationary point ππ¦
ππ₯ = 0
-12x(5 β 2x2)2 = 0 M1
x = 0
f(0) = (5 β 2(0)2)3 = (125) M1
x = Β±1
2β10
f(+1
2β10) = 0
M1
f(-1
2β10) = 0 M1
(-1
2β10, 0)
(0, 125)
(1
2β10, 0)
10c.
x = 3
2
f(3
2) = (5 β 2(
3
2)2)3 =
1
8
M1
Gradient = -18 x 1
4 = -
9
2 M1
8y β 1 = -36x + 54
36x + 8y β 55 = 0 M1
11a.
fβ(x) = 3
π₯ - 2 M1
11b.
Gradient of curve = 4 M1 3
π₯ β 2 = 4
3
π₯ = 6
x = 1
2
M1
11c.
Stationary point, fβ(x) = 0 M1 3
π₯ β 2 = 0
x = 3
2
M1
f(3
2) = 3 ln 5(
3
2) β 2(
3
2)
f(3
2) = 3 ln
15
2β 3
M1
(3
2, 3 ln
15
2β 3)
11d.
x β₯ 3
2 M1
12a.
fβ(x) = 2x β 7 + 4
π₯ = 0 M1
2x2 β 7x + 4 = 0
x = 7Β± β17
4
M1
x = 0.72
x = 2.78 M1
12b.
x = 2
Therefore, y = -10
gradient = -1
M1
y + 10 = -(x-2)
y = -x - 8 M1
Differentiation
Pt. 5: Product Rule
1. Differentiate xex with respect to x. (3)
2. Find ππ¦
ππ₯ when y = x2 ln (x + 6) (3)
3. Find the coordinates of any stationary points of the curve y = xβπ₯ + 12 (4)
4. Find an equation for the tangent to the curve y = (4x β 1)ln2x (5)
5. Find an equation for the normal to the curve y = (x2 β 1)e3x (5)
6. The diagrams shows part of the curve with equation y = xππ₯2 and the tangent to the curve at the point P with x-
coordinate 1.
a. Find an equation for the tangent to the curve at P. (5)
b. Show that the area of the triangle bounded by this tangent and the coordinate axes is 2
3π. (3)
7. A curve has the equation y = eβx sin x.
a. Find ππ¦
ππ₯ and
π2π¦
ππ₯2 (4)
b. Find the exact coordinates of the stationary points of the curve in the interval -π β€ π₯ β€ π and determine their
nature. (8)
8. The diagram shows the curve y = f(x) in the interval 0 β€ π₯ β€ 2π, where f(x) = cos x sin 2x
a. Show that fβ(x) = 2 cos x(1 β 3sin2x) (2)
b. Find the x-coordinates of the stationary points of the curve in the interval 0 β€ π₯ β€ 2π (4)
c. Show that the maximum value of f(x) in the interval 0 β€ π₯ β€ 2π is 4
9β3 (4)
d. Explain why this is the maximum value of f(x) for all real values of x. (2)
A-Level
Pt. 3: General Differentiation
Pt. 4: The Chain Rule
Pt. 5: The Product Rule
Pt. 6: The Quotient Rule
Pt. 7: Parametric Differentiation
Pt. 8: Implicit Differentiation
Pt. 9: Rates of Change
AS Level
Pt. 1: First Principles
Pt. 2: Differentiation
Pt. 9: Rates of Change
Pt. 1: First Principles
Pt. 2: Differentiation
Mark Scheme
1.
u = x ππ’
ππ₯ = 1
v = ex ππ£
ππ₯ = ex
M1
π
ππ₯(π₯πx) = 1 Γ ex + x Γ ex M1
π
ππ₯(π₯πx) = ex(1 + x) M1
2.
u = x2 ππ’
ππ₯ = 2x
v = ln (x + 6) ππ£
ππ₯ =
1
π₯+6
M1
ππ¦
ππ₯ = 2x x ln(x + 6) + x2 x
1
π₯+6 M1
ππ¦
ππ₯ = 2x ln (x + 6) +
π₯2
π₯+6 M1
3.
u = x ππ’
ππ₯ = 1
v = βπ₯ + 12 = (x + 12)0.5 ππ£
ππ₯ = 0.5(x + 12)-0.5
M1
ππ¦
ππ₯ = 1 Γ βπ₯ + 12 + x Γ 0.5(x + 12)-0.5
ππ¦
ππ₯ = 0.5(x + 12)-0.5 [2(x + 12) + x]
M1
At stationary point ππ¦
ππ₯ = 0,
3(π₯ + 8)
2βπ₯ + 12= 0
x = -8
M1
When x = -8,
y(-8) = (-8)β(β8) + 12 = 16 M1
(-8, -16)
4.
x = 1
2
y(1
2) = 0
M1
u = 4x - 1 ππ’
ππ₯ = 4
v = ln 2π₯ ππ£
ππ₯ =
1
π₯
ππ¦
ππ₯= 4 Γ ππ2π₯ + (4π₯ β 1) Γ
1
π₯
M1
ππ¦
ππ₯ = 4 ln 2x + 4 -
1
π₯ M1
ππ¦
ππ₯ (
1
2) = 4 ln 2(
1
2) + 4 -
11
2
= 2 M1
y β 0 = 2(x - 1
2)
y = 2x - 1 M1
5.
When x = 0, y = -1 M1
u = (x2 β 1) ππ’
ππ₯ = 2x
v = e3x ππ£
ππ₯ = 3e3x
ππ¦
ππ₯= 2π₯ Γ π3π₯ + (π₯2 β 1) Γ 3π3π₯
M1
ππ¦
ππ₯ = π3π₯[2x + 3(π₯2 β 1)]
= π3π₯(3x2 + 2x β 3) M1
ππ¦
ππ₯ (0) = -3
Therefore gradient of normal = 1
3
M1
y + 1 = 1
3 (x β 0)
x β 3y β 3 = 0 M1
6a.
When x = 1
y (1) = xππ₯2= (1)π(1)2
= e M1
u = x ππ’
ππ₯ = 1
v = ππ₯2
ππ£
ππ₯ = 2xππ₯2
ππ¦
ππ₯= 1 Γ ππ₯2
+ π₯ Γ ππ₯2 Γ 2π₯
M1
ππ¦
ππ₯ = ππ₯2
(1 + 2π₯2) M1
ππ¦
ππ₯ (1) = π(1)2
(1 + 2(1)2) = 3e M1
y β e = 3e(x β 1)
y = e(3x β 2) M1
6b.
When x = 0
y = -2e M1
When y = 0,
x = 2
3 M1
Therefore area = 1
2 Γ 2π Γ
2
3=
2
3π M1
7a.
u = e-x
ππ’
ππ₯ = -e-x
v = sin x ππ£
ππ₯ = cos x
ππ¦
ππ₯= -e-x Γ sin π₯ + π-x Γ cos x
M1
ππ¦
ππ₯ = e-x(cos x β sin x) M1
π
ππ₯ e-x(cos x β sin x)
u = e-x
ππ’
ππ₯ = -e-x
v = cos x β sin x ππ£
ππ₯ = - sin x β cos x
π2π¦
ππ₯2 = -e-x Γ (cos x β sin x) + e-x Γ (- sin x β cos x)
M1
π2π¦
ππ₯2 = β2e-x cos x M1
7b.
At stationary point, ππ¦
ππ₯ = 0
e-x(cos x β sin x) = 0 M1
cos x β sin x = 0
tan x = 1 M1
x = π
4
π2π¦
ππ₯2 = ββ2 πβπ
4
π2π¦
ππ₯2 < 0 , therefore maximum point
M1
M1
x = β3π
4
π2π¦
ππ₯2 = β2 π3π
4
π2π¦
ππ₯2 > 0 , therefore minimum point
M1
M1
When x = π
4, y =
1
β2πβ
π
4 , maximum point M1
When x = β3π
4, y = β
1
β2π
3π
4 , minimum point M1
8a.
fβ(x) = - sin x Γ sin 2x + cos x Γ 2cos 2x M1
fβ(x) = 2cos x(1 β 2sin2x) β 2sin2x cos x
= 2 cos x(1 β 3 sin2x) M1
8b.
At stationary point fβ(x) = 0
2 cos x(1 β 3 sin2x) = 0 M1
Cos x = 0
sin x = = Β±1
β3
M1
x = 0.615, π
2, 2.53, 3.76,
3π
2, 5.76 M1 M1
8c.
f(x) at max when sin x = 1
β3 M1
cos2x = 1 β sin2x = 1 β 1
3 =
2
3 M1
f(x) = cos x Γ 2 sin π₯ cos π₯ = 2 sin x cos2x
M1
Therefore maximum = 2 Γ1
β3 Γ
2
3 =
4
9β3 M1
8d.
Period of cos x = 2π
Period of sin x = π M1
Therefore period of f(x) = 2π
Values of f(x) in this interval are repeated. M1
Differentiation
Pt. 6: Quotient Rule
1. Differentiate ππ₯
π₯β4 with respect to xand simplify your answer (2)
2. Find ππ¦
ππ₯ of y =
ln(3π₯β1)
π₯+2 (2)
3. Find the co-ordinates of any stationary point on the curve y = π₯+5
β2π₯ +1 (4)
4. Find the equation of the tangent to the curve y = 3π₯+4
π₯2+1 at the point when x = -1 (5)
5. The curve C has the equation y = 3+sin 2π₯
2+cos 2π₯
a. Show that ππ¦
ππ₯=
6 sin 2π₯+4 cos 2π₯+2
(2+cos 2π₯)2 (4)
b. Find an equaton of the tangent to C at the point where x = π
2. Write your answer in the form y = ax+ b, where a
and b are exact constants. (4)
6. f(x) = π₯β5
2π₯+3+
2π₯+4
2π₯2+7π₯+6
a. Express f(x) as a fraction in its simplest form (2)
b. Hence find fβ(x) in its simplest form (3)
7. Differentiate cos 2π₯
βπ₯ with respect to x (3)
8. Given that y = 3 sin π₯
2 sin π₯+2 cos π₯, show that
ππ¦
ππ₯=
π΄
1+sin 2π₯ where A is a rational constant to be found. The value of xlies
in the domain βπ
4 < π₯ <
3π
4 (5)
9. Find equation of the normal to the curve y = ββπ₯
5 cos 5π₯ at the x-cordiante 16. (7)
A-Level
Pt. 3: General Differentiation
Pt. 4: The Chain Rule
Pt. 5: The Product Rule
Pt. 6: The Quotient Rule
Pt. 7: Parametric Differentiation
Pt. 8: Implicit Differentiation
Pt. 9: Rates of Change
AS Level
Pt. 1: First Principles
Pt. 2: Differentiation
Pt. 9: Rates of Change
Pt. 1: First Principles
Pt. 2: Differentiation
Mark Scheme
1. π
ππ₯(
ππ₯
π₯β4 )
= ππ₯ Γ(π₯β4)β ππ₯ Γ1
(π₯β4)2 M1
= ππ₯ (π₯β5)
(π₯β4)2 M1
2. π
ππ₯(
ln(3π₯β1)
π₯+2) )
=
3
3π₯β1Γ(π₯+2)βln (3π₯β1)Γ1
(π₯+2)2 M1
= 3
(3π₯β1)(π₯+2)β
ln (3π₯β1)
(π₯+2)2 M1
3.
ππ¦
ππ₯=
1Γβ2π₯+1β(π₯+5)Γ1
2(2π₯+1)
12Γ2
2π₯+1 M1
ππ¦
ππ₯=
(2π₯+1)β(π₯+5)
(2π₯+1)32
=π₯β4
(2π₯+1)32
M1
At stationary points, ππ¦
ππ₯ = 0
π₯β4
(2π₯+1)32
= 0
xβ 4 = 0
x= 4
M1
When x= 4
y(4) = 4+5
β2(4) +1 = 3 M1
Stationary point = (4, 3)
4. ππ¦
ππ₯=
3 Γ(π₯2+1)β(3π₯+4)Γ2π₯
(π₯2+1)2 M1
ππ¦
ππ₯ =
3β8π₯β3π₯2
(π₯2+1)2 M1
ππ¦
ππ₯(β1) = 2 M1
When x= -1
y(-1) = 1
2
M1
y β 1
2= 2(π₯ + 1) =
y = 2x+ 5
2
M1
5a. ππ¦
ππ₯=
(2+cos 2π₯)(2 cos 2π₯)β(3+sin 2π₯)(β2π ππ2π₯)
(2+πππ 2π₯)2 M1
ππ¦
ππ₯ =
4 cos 2π₯ + 2πππ 22π₯ + 6π ππ2π₯ + 2π ππ22π₯
(2+πππ 2π₯)2 M1
ππ¦
ππ₯=
6 sin 2π₯ + 4 cos 2π₯ + 2(πππ 22π₯ + π ππ22π₯)
(2+πππ 2π₯)2 M1
ππ¦
ππ₯ =
6 sin 2π₯ + 4 cos 2π₯ + 2
(2+πππ 2π₯)2 M1
5b.
When x= π
2
ππ¦
ππ₯ =
6 sin 2(π
2) + 4 cos 2(
π
2) + 2
(2+πππ 2(π
2))2
= -2 M1
When x= π
2
y(π
2) =
3+sin 2(π
2)
2+cos 2(π
2) = 3
M1
y β 3 = -2(x- π
2) M1
y = -2x+ π + 3 M1
6a.
f(x) = π₯β5
2π₯+3+
2(π₯+2)
(2π₯+3)(π₯+2) M1
= π₯β5
2π₯+3+
2
2π₯+3
= π₯β3
2π₯+3
M1
6b.
fβ(x) = 2π₯+3β2(π₯β3)
(2π₯+3)2 M1
M1
fβ(x) = 2π₯+3β2π₯+6
(2π₯+3)2
fβ(x) = 9
(2π₯+3)2 M1
7.
u = cos 2x ππ’
ππ₯= β2 sin 2π₯
v = π₯1
2 ππ£
ππ₯ =
1
2π₯β
1
2
M1
π
ππ₯(
π’
π£) =
π₯12(β2 sin 2π₯)βπππ 2π₯(
1
2π₯
β12)
(π₯12)2
M1
π
ππ₯(
π’
π£) =
β2π₯12 sin 2π₯ β
1
2π₯
β12 cos 2π₯
π₯ M1
8.
u = 3 sin x ππ’
ππ₯= 3 cos π₯
v = 2 sin x+ 2 cos x ππ£
ππ₯ = 2 cos xβ 2 sin x
M1
π
ππ₯(
π’
π£) =
3 cos π₯(2 sin π₯+2 cos π₯)β(2 cos π₯β2 sin π₯)(3 sin π₯)
(2 sin π₯+2 cos π₯)2 M1
= 6
4 Γ
cos π₯ sin π₯ + πππ 2π₯βcos π₯ sin π₯+ π ππ2π₯
π ππ2π₯+2 sin π₯ cos π₯+ πππ 2π₯
USING:
(π ππ2π₯ + πππ 2π₯ = 1)
(sin 2x= 2 sin xcos x)
M1
M1
π
ππ₯(
π’
π£) =
3
2
1+sin 2π₯ M1
9.
u = -βπ₯ = -x0.5
ππ’
ππ₯= β0.5 π₯-0.5
v = 5cos 5x ππ£
ππ₯ = -5 sin 5xΓ 5 = -25 sin 5x
M1
π
ππ₯(
π’
π£) =
(5 cos 5π₯) (β0.5π₯β0.5)β(βπ₯0.5)(β25 sin 5π₯)
(5 cos 5π₯)2 M1
At the point x= 1 , ππ¦
ππ₯=
(5 cos 5) (β0.5β0.5)β(β10.5)(β25 sin 5)
(5 cos 5)2
ππ¦
ππ₯ = 10.9β¦
M1
M1
Gradient of normal = -0.1 M1
When x= 1, y = ββ1
5 cos 5 = -0.71 M1
y = mx+ c
-0.71 = (-0.1)(1) + c
c = -0.61
M1
y = -0.1xβ 0.61
Differentiation
Pt. 7: Parametric Differentiation
1. A curve is given by the parametric equations x = t2 + 1 and y = 4
π‘
a. Write down the coorindates of the point on the curve when t = 2 (2)
b. Find the value of t at the point on the curve with coordinates (5
4, β8) (1)
2. Find the cartesian equation for the curve, given its parametric equations are x = 3t and y = t2 (2)
3. Find the cartesian equations for the curve, given its parametric equations are x = 2t β 1 and y = 2
π‘2 (2)
4. Write down the parametic equation for a circle with centre (6, -1) and radius 2. (2)
5. A curve is given by the parametric equations x = 2 + t and y = t2 β 1
a. Write down expression for ππ₯
ππ‘ and
ππ¦
ππ‘ (1)
b. Hence, show that ππ¦
ππ₯= 2π‘ (1)
6. Find and simplify an expression for ππ¦
ππ₯ in terms of the parameter t when x = 3t β 1 and y = 2 -
1
π‘ (3)
7. Find and simplify an expression for ππ¦
ππ₯ in terms of the parameter t when x = sin2t and y = cos3t (3)
8. A curve has the parametric equations x = t3 and y = 3t2. Find, in the form y = mx + c, an equation for the tangent
to the curve at the point with the given value of the parameter t. (4)
9. Show that the normal to the curve with parametric equations, x = sec π, y = tan π, 0 β€ π < π
2 at the point where
π = π
3, has the equation β3π₯ + 4π¦ = 10β3 (6)
10. A curve has parametric equations x = sin 2t and y = sin2t, 0 < t < π
a. Show that ππ¦
ππ₯ =
1
2tan 2π‘ (2)
b. Find an equation for the tangent to the curve at the point where t = π
6 (3)
11. A curve is given by the parametric equations
x = t 2, y = t(t β 2), t β₯ 0.
a. Find the coordinates of any points where the curve meets the coordinate axes. (3)
b. Find ππ¦
ππ₯ in terms of x by first finding
ππ¦
ππ₯ in terms of t. (3)
c. Find ππ¦
ππ₯ in terms of x by first finding a cartesian equation for the curve. (2)
12. A curve has parametric equations x = sin2t and y = tan t, βπ
2 < t <
π
2
a. Show that the tangent to the curve at the point where t = π
4 passes through the origin. (6)
b. Find a cartesian equation for the curve in the form y2 = f(x). (2)
A-Level
Pt. 3: General Differentiation
Pt. 4: The Chain Rule
Pt. 5: The Product Rule
Pt. 6: The Quotient Rule
Pt. 7: Parametric Differentiation
Pt. 8: Implicit Differentiation
Pt. 9: Rates of Change
AS Level
Pt. 1: First Principles
Pt. 2: Differentiation
Pt. 9: Rates of Change
Pt. 1: First Principles
Pt. 2: Differentiation
Mark Scheme
1a.
x(2) = 22 + 1 = 5 M1
y(2) = 4
2 = 2 M1
1b. 4
π‘= β8
t = -1
2
M1
2.
x = 3t β t = π₯
3 M1
y = (π₯
3)2 =
π₯2
9 M1
3.
t = 1
2(x + 1) M1
y = 2
1
4(π₯+1)2
=8
(π₯+1)2 M1
4.
x = 6 + 2 cos x M1
y = -1 + 2 sin x M1
0 β€ x < 2π
5a. ππ₯
ππ‘= 1
ππ¦
ππ‘= 2π‘
M1
5b. ππ¦
ππ₯=
ππ¦
ππ‘ Γ·
ππ₯
ππ‘
ππ¦
ππ₯ = 2t Γ· 1 = 2t
M1
6.
ππ₯
ππ‘= 3
ππ¦
ππ‘= π‘β2
M1
ππ¦
ππ₯=
ππ¦
ππ‘ Γ·
ππ₯
ππ‘
ππ¦
ππ₯ = t-2 Γ· 3 =
π‘β2
3
M1
ππ¦
ππ₯ =
1
3π‘2 M1
7.
ππ₯
ππ‘= 2 sin π‘ Γ cos π‘
ππ¦
ππ‘= 3cos2t Γ (- sin t)
M1
ππ¦
ππ₯=
ππ¦
ππ‘ Γ·
ππ₯
ππ‘
= β3πππ 2π‘ sin π‘
2 sin π‘ cos π‘
M1
ππ¦
ππ₯ = -
3
2cos π‘ M1
8.
when t = 1
x(1) = 1
y(1) = 3
M1
ππ₯
ππ‘= 3t2
ππ¦
ππ‘= 6t
M1
ππ¦
ππ₯ =
6π‘
3π‘2 ππ¦
ππ₯ (1) = 2
M1
y β 3 = 2(x β 1)
y = 2x + 1 M1
9.
π = π
3
x(π
3) = 2
y(π
3) = 2β3
M1
ππ₯
ππ = sec π tan π
ππ¦
ππ= 2sec2 π
M1
ππ¦
ππ₯ =
2π ππ2π
sec π tan π
ππ¦
ππ₯ =
2 sec π
tan π= 2 πππ ππ π
M1
Gradient = 4
β3
Therefore gradient of normal = -β3
4
M1
y - 2β3 = -β3
4(x β 2)
4y - 8β3 = -β3x + 2β3 M1
β3x + 4y = 10β3 M1
10a. ππ₯
ππ‘= 2 cos 2π‘
ππ¦
ππ‘= 2 sin π‘ Γ cos π‘ = sin 2π‘
M1
ππ¦
ππ₯=
sin 2π‘
2 cos 2π‘=
1
2tan 2π‘ M1
10b.
t = π
6
x(π
6) =
β3
2
y(π
6) =
1
4
M1
Gradient = β3
2
y - 1
4 =
β3
2(x -
β3
2)
M1
β3 x β 2y β 1 = 0 M1
11a.
x = 0
t2 = 0
t = 0
M1
y = 0
t(t β 2) = 0
t = 0 or t = 2
M1
Co-ordinates: (0, 0) and (4, 0) M1
11b.
ππ₯
ππ‘ = 2t
ππ¦
ππ‘= 2π‘ β 2
M1
ππ¦
ππ₯=
2π‘β2
2π‘= 1 β t -1 M1
t = x0.5 ππ¦
ππ₯ = 1 β x-0.5 M1
11c.
y = π₯1
2(π₯1
2 β 2) = x - 2π₯1
2 M1
ππ¦
ππ₯= 1 β π₯β
1
2 M1
12a.
t = π
4
x(π
4) =
1
2
y(π
4) = 1
M1
ππ₯
ππ‘ = 2 sin t cos t
ππ¦
ππ‘= sec2 t
M1
ππ¦
ππ₯=
π ππ2π‘
2 sin π‘ cos π‘=
1
2 π ππ3π‘ πππ ππ π‘ M1
Gradient = 1
2Γ (β2)3 Γ β2 = 2 M1
y β 1 = 2(x β 1
2)
y = 2x M1
When x = 0, y = 0 therefore passes through origin M1
12b.
y2 = tan2t = π ππ2π‘
πππ 2π‘ =
π ππ2π‘
1β π ππ2π‘ M1
y2 = π₯
1βπ₯ M1
Differentiation
Pt. 8: Implicit Differentiation
1. Differentiate with respect to x, y3 (1)
2. Find ππ¦
ππ₯ in terms of x and y for the curve x2 + y2 + 3x β 4y = 9 (2)
3. Find ππ¦
ππ₯ in terms of x and y for the curve 2e3x + e-2y + 7 = 0 (3)
4. Find ππ¦
ππ₯ in terms of x and y for the curve 2xy2 β x3y = 0 (4)
5. Find ππ¦
ππ₯ in terms of x and y for the curve cos 2x sec 3y + 1 = 0 (3)
6. A curve has the equation x2 + 4xy β 3y2 = 36.
a. Find an equation for the tangent to the curve at the point P (4, 2). (5)
Given that the tangent to the curve at the point Q on the curve is parallel to the tangent at P,
b. Find the coordinates of Q. (4)
7. A curve has the equation x2 β 4xy + y2 = 24.
a. Show that ππ¦
ππ₯=
π₯β2π¦
2π₯βπ¦ (3)
b. Find an equation for the tangent to the curve at the point P(2, 10) (2)
The The tangent to the curve at Q is prallallel to the tangent at P
c. Find the coordinates of Q. (5)
A-Level
Pt. 3: General Differentiation
Pt. 4: The Chain Rule
Pt. 5: The Product Rule
Pt. 6: The Quotient Rule
Pt. 7: Parametric Differentiation
Pt. 8: Implicit Differentiation
Pt. 9: Rates of Change
AS Level
Pt. 1: First Principles
Pt. 2: Differentiation
Pt. 9: Rates of Change
Pt. 1: First Principles
Pt. 2: Differentiation
Mark Scheme
1. π
ππ₯(π¦3) = 3y2 ππ¦
ππ₯ M1
2. π
ππ₯(x2 + y2 + 3x β 4y = 9) β 2x + 2y
ππ¦
ππ₯ + 3 - 4
ππ¦
ππ₯ = 0 M1
ππ¦
ππ₯ (4 β 2y) = 2x + 3
ππ¦
ππ₯ =
2π₯+3
4β2π¦
M1
3. π
ππ₯(2e3x + e-2y + 7 = 0) β 6e3x β 2e-2y
ππ¦
ππ₯ = 0 M1
6e3x = 2e-2y ππ¦
ππ₯ M1
ππ¦
ππ₯ =
3π3π₯
πβ2π¦ = 3e3x + 2y M1
4. π
ππ₯(2xy2) β product rule
u = 2x
uβ = 2
v = y2
vβ = 2y ππ¦
ππ₯
π
ππ₯(2xy2) β (2 Γ y2) + (2x Γ 2y
ππ¦
ππ₯)
M1
π
ππ₯(π₯3y) β product rule
u = x3
uβ = 3x2
v = y
vβ = ππ¦
ππ₯
π
ππ₯(π₯3y) β (3x2 Γ y) + (x3 Γ
ππ¦
ππ₯)
M1
π
ππ₯(2xy2 β x3y) = (2 Γ y2) + (2x Γ 2y
ππ¦
ππ₯) β ((3x2 Γ y) + (x3 Γ
ππ¦
ππ₯))
2y2 β 3x2y = ππ¦
ππ₯(x3 β 4xy)
M1
ππ¦
ππ₯=
2π¦2β3π₯2π¦
π₯3β4π₯π¦ M1
5. π
ππ₯(cos 2x sec 3y + 1) β
π
ππ₯(cos 2x sec 3y) + 0 β Product Rule
u = cos 2x
uβ = -2sin 2x
v = sec 3y
vβ = 3ππ¦
ππ₯ sec 3y tan 3y
π
ππ₯(cos 2x sec 3y) β (-2 sin 2x Γ sec 3y + cos 2x Γ 3
ππ¦
ππ₯ sec 3y tan 3y)
M1
3ππ¦
ππ₯ cos 2x sec 3y tan 3y = 2sin 2x sec 3y M1
ππ¦
ππ₯ =
2 sin 2π₯
3 cos 2π₯ tan 3π¦=
2
3tan 2π₯ cot 3π¦ M1
6a. π
ππ₯ (x2 + 4xy β 3y2 = 36)
2x + 4 Γ y + 4x Γ ππ¦
ππ₯ β 6y
ππ¦
ππ₯ = 0
M1
2x + 4y = ππ¦
ππ₯(6π¦ β 4π₯) M1
ππ¦
ππ₯=
π₯+2π¦
3π¦β2π₯ M1
ππ¦
ππ₯(4, 2) = β4 M1
y β 2 = -4(x β 4)
y = 18 β 4x M1
6b.
At Q, π₯+2π¦
3π¦β2π₯ = -4
x + 2y = -4(3y β 2x)
x = 2y
M1
Sub into equation of the curve:
(2y)2 + 4y(2y) β 3y2 = 36 M1
y2 = 4
y = 2 or -2 M1
Therefore Q: (-4, -2) M1
7a. π
ππ₯ (x2 β 4xy + y2 = 24) β 2x β 4 Γ y β 4x Γ
ππ¦
ππ₯ + 2y
ππ¦
ππ₯ = 0 M1
2x β 4y = ππ¦
ππ₯(4x β 2y) M1
ππ¦
ππ₯ =
2π₯β4π¦
4π₯β2π¦=
π₯ β 2π¦
2π₯ β π¦ M1
7b.
Gradient = 3 M1
y β 10 = 3(x β 2)
y = 3x + 4 M1
7c. π₯ β 2π¦
2π₯ β π¦= 3 M1
x β 2y = 3(2x β y)
y = 5x M1
x2 β (4x)(5x) + (5x)2 = 24 M1
x2 = 4
x = or x = -2 M1
Therefore, (-2, -10) M1
Differentiation
Pt. 9: Rates of Change
1. A biological culture is growing exponentally such that the number of bacteria present, N, at the time t minutes is
given by N = 8000(1.04)t. Find the rate at which the number of bacteria is increasing when there are 4000 bacteria
present. (3)
2. The radius of a circle is increasing at a constant rate of 0.2 cm sβ1.
a. Show that the perimeter of the circle is increasing at the rate of 0.4Ο cm sβ1. (3)
b. Find the rate at which the area of the circle is increasing when the radius is 10 cm. (3)
c. Find the radius of the circle when its area is increasing at the rate of 20 cm2 sβ1. (2)
3. The volume of a cube is increasing at a constant rate of 3.5 cm3 s-1. Find,
a. The rate at which the length of one side of the cube is increasing when the volume is 200 cm3 (5)
b. The volume of the cube when the length of one side is increasing at the rate of 2 mmsβ1 (4)
4. The diagram shows the cross-section of a right-circular paper cone being used as a filter funnel. The volume of
liquid in the funnel is V cm3 when the depth of the liquid is h cm. Given that the angle between the sides of the
funnel in the cross-section is 60Β° as shown.
a. Show that V = 1
9πh3 (2)
Given also that at time t seconds after liquid is put in the funnel, V = 600eβ0.0005t
b. Show that after two minutes, the depth of liquid in the funnel is approximately 11.7 cm (2)
c. Find the rate at which the depth of liquid is decreasing after two minutes. (6)
5. The volume, V m3, of liquid in a container is given by: V = (3h2 + 4)3
2 β 8 where h m is the depth of the liquid.
a. Find the value of ππ
πβ when h = 0.6, giving your answer correct to 2 decimal places. (4)
b. Liquid is leaking from the container. It is oberseved that, when the depth of the liquid is 0.6m, the depth is
decreasing at a rate of 0.015m per hour. Find the rate at which the volume in the containiner is decreasing at the
instant when the depth is 0.6m (3)
A-Level
Pt. 3: General Differentiation
Pt. 4: The Chain Rule
Pt. 5: The Product Rule
Pt. 6: The Quotient Rule
Pt. 7: Parametric Differentiation
Pt. 8: Implicit Differentiation
Pt. 9: Rates of Change
AS Level
Pt. 1: First Principles
Pt. 2: Differentiation
Pt. 9: Rates of Change
Pt. 1: First Principles
Pt. 2: Differentiation
Mark Scheme
1. ππ
ππ‘= 800(1.04)t x ln 1.04 M1
N = 4000
4000 = 800(1.04)t
(1.04)t = 5
M1
ππ
ππ‘= 800 Γ 5 Γ ln 1.04 = 157
Therefore, growing at a rate of 157 per minute. M1
2a. ππ
ππ‘=
ππ
ππΓ
ππ
ππ‘
ππ
ππ‘= 0.2
M1
P = 2ππ ππ
ππ= 2π
M1
ππ
ππ‘= 2π Γ 0.2 = 0.4π
Therefore perimeter is increasing at 0.4π cm s-1 M1
2b. ππ΄
ππ‘=
ππ΄
ππΓ
ππ
ππ‘
A = πr2 ππ΄
ππ= 2ππ
M1
When r = 10, ππ΄
ππ= 20π M1
ππ΄
ππ‘= 20π Γ 0.2 = 4π cm2 s-1 M1
2b.
20 = 2ππ Γ 0.2 M1
r = 50
π = 15.9 cm M1
3a. ππ
ππ‘=
ππ
ππ Γ
ππ
ππ‘
ππ
ππ‘= 3.5, V = l3
ππ
ππ= 3l2
M1
200 = l3
l = 5.848 M1
ππ
ππ= 3 Γ 5.8482 = 102.6 M1
3.5 = 102.6 x ππ
ππ‘ M1
ππ
ππ‘= 3.5 Γ· 102.6 = 0.0341 ππs-1 M1
3b.
2 mm s-1 = 0.2 cm s-1 M1
3.5 = ππ
ππ Γ 0.2
ππ
ππ = 17.5
M1
17.5 = 3l2
l = 2.415 M1
V = l3 = 14.1 cm3 M1
4a.
tan 30 = 1
β3 =
π
β
r = β
β3
M1
V = 1
3πr2h =
1
3π Γ
1
3β2 Γ β =
1
9πβ3 M1
4b.
t = 120, V = 565.06
h = β9 Γ565.06
π
3
M1
h = 11.7cm M1
4c. h = 11.74 ππ
πβ= 144.4
M1
ππ
ππ‘=
ππ
πβΓ
πβ
ππ‘
ππ
πβ=
1
3πh2
M1
ππ
ππ‘= 600 Γ (β0.0005)e-0.0005t
= -0.3e-0.0005t M1
t = 120 ππ
ππ‘= β0.2825
M1
0.2825 = 144.4 x πβ
ππ‘
πβ
ππ‘= β0.00196
M1
The depth is decreasing at 0.00196 cm s-1 M1
5a. ππ
πβ=
3
2(3β2 + 4)
1
2(6β) = 3(3β2 + 4)1
2(3β) M1
= 9h(3h2 + 4)1
2 M1
When h = 0.6, ππ
πβ= 9(0.6)(3(0.62) + 4)
12 M1
ππ
πβ= 12.17 M1
5b. ππ
ππ‘=
ππ
πβΓ
πβ
ππ‘ M1
When h = 0.6, ππ
ππ‘ = (12.170) x (-0.015) M1
ππ
ππ‘ = -0.1825
The volume is decreasing at a rate of 0.18m3 per hour. M1