Download - Topic 7
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Topic 7
Rates of Change II
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Rates of ChangeRules for differentiation including
ruleChain)]([
ruleProduct)]()([
)]()([
)]([
nofvaluesrationalfor
xgf
tgtf
sgsf
rkf
p
dxd
dtd
dsd
drd
ndpd
evaluation of the derivative of a function at a point (SLE 3,4)
interpretation of the derivative as the instantaneous rate of change (SLE 1,2,5,10)
interpretation of the derivative as the gradient function (SLE 1,2,4,6)
practical applications of instantaneous rates of change (SLE 1-5,7-11)
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Derivative of y = axn
When we did differentiation from first principles, we found that if
xthenxy dxdy 22
23 3xthenxy dxdy
34 4xthenxy dxdy
45 5xthenxy dxdy
Are you seeing a pattern here?
1 ndxdyn nxthenxy
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Derivative of y = axn
When we did differentiation from first principles, we found that if
xthenxy dxdy 84 2
23 155 xthenxy dxdy
34 82 xthenxy dxdy
45 204 xthenxy dxdy
Can you see what’s happening?
1 ndxdyn anxthenaxy
Calculus Phobe Videos
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Model: Find the derivative of y = 3x4 – 2x3 + 5x2 + 7x + 8
y = 3x4 – 2x3 + 5x2 + 7x + 8 = 12x3 – 6x2 + 10x + 7
dy
dx
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Model: Find the derivative of y = 3x2 + 2x -5 at x = (a) 1 (b) -1
y = 3x2 + 2x - 5 = 6x + 2
(a) When x = 1, = 6x1 + 2 = 8
(b) When x = -1, = 6x-1 + 2 = -4
dy
dx
dy
dx
dy
dx
X
Y
-2 2
-10-8
-6-4
-2
2
46
810
0
y = 3x2+2x-5
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Gradient is the same as derivative
When you hear “gradient”, you
think
When you hear
you think “gradient
”
dx
dy
dx
dy
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Exercise
NewQ
Set 8.2 Page 262 No. 2 Set 8.3 Page 266 No.4,7-9
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Model Differentiate each of the following:
22
2
)63()(
3)(
3)(
xxyc
xyb
xya
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3
3
2
2
6
32
3
3)(
x
xdx
dy
x
xya
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x
x
x
xdx
dy
x
xyb
2
32
32
3
3
3
3)(
2
1
21
2
1
2
1
2
1
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366184
363636
)63)(63(
)63()(
23
234
22
22
xxxdx
dy
xxxx
xxxx
xxyc
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Exercise
NewQ
Set 8.3 Page 266 No.1Set 8.4 Page 270 No. 1-2
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The Chain Rule
dx
du
du
dy
dx
dy
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366184
363636
)63)(63(
)63()(
23
234
22
22
xxxdx
dy
xxxx
xxxx
xxyc
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To be used to differentiate a function of a function
Recall our last function
dy dy du
dx du dx
366184
363636
)63)(63(
)63()(
23
234
22
22
xxxdx
dy
xxxx
xxxx
xxycLet u = x2 + 3x – 6 ∴ y = u2
)63)(64(
)64(
)32(2
232
2
xxx
ux
xudx
du
du
dy
dx
dy
udu
dyx
dx
du
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Try these…a) y = (3x2 + 5x – 2)2 b) y = (5x3 – 6x2 + 3x +
5)3
a) (12x + 10)(3x2 + 5x – 2)
b) (45x2 – 36x + 9)(5x3 – 6x2 + 3x + 5)2
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Exercise
NewQ
Set 8.4 Page 270 No. 5(a&b), 6(i-t)
Set 8.5 Page 273 No. 1(a,e,f)
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The Product Rule
dx
duv
dx
dvu
dx
dy
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To be used when differentiating a function multiplied by a function
dx
duv
dx
dvu
dx
dy
If y = (3x + 4) (5x2 – 2x + 1)
Let u = 3x+ 4 and v = 5x2 – 2x + 1
210 and3 xdx
dv
dx
du
52845
361583430
3)125()210)(43(
2
22
2
xx
xxxx
xxxx
dx
duv
dx
dvu
dx
dy
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Exercise
NewQ
Set 8.5 Page 273 No. 2-3
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Answers 8.5 number 3
a) 15(3x-2)4(x+8)2 + 60(3x-2)3(x+8)3
b) 4(4t-5)3(2t-7) + 12(4t-5)2(2t-7)2
c) 48(m-6)(4m+3)7 + 6(4m+3)8
d) 21(2k+9)4(3k-5)6 + 8(2k+9)3(3k-5)7
e) 24(3g-4)5(5g-4)5 + 60(3g-4)4(5g-4)6
f) 112(12-5t)9(4t-11)3 + 315(12-5t)8(4t-11)4
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The Quotient Rule
2vdxdvu
dxduv
dx
dy
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16249
3016249
)618()2418(
)43(
3)26(6)43(
2
2
2
xx
xx
xx
x
xx
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Exercise
Set 8.5 Page 273 No. 4-5
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Application of Derivatives
f(x) = 5x – 3x2 + 8f’(x) = 5 – 6xf’(2) = 5 – 6 × 2m = -7
At x = 2, y = 6, m = -7
∴ y = mx + c 6 = -7 × 2 + c c = 20 y = -7x + 20
Model No. 1
If the equation of a parabola is y = 5x – 3x2 + 8, at the point where x = 2, find:a) The gradient of the tangent b) the equation of the tangent
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Model No. 2
The volume of liquid in an underground tank being gravity filled from a tanker is given by V = 1200 + 1020t – 17t2 where the volume, V, is in litres and time, t, is in minutes.
a) What is the filling rate after 5 minutes?
b) What is the flow rate in the pipe after 10 minutes?
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The volume of liquid in an underground tank being gravity filled from a tanker is given by
V = 1200 + 1020t – 17t2 where the volume, V, is in litres and time, t, is in minutes.
a) What is the filling rate after 5 minutes?
b) What is the flow rate in the pipe after 10 minutes?
Filling rate and flow rate in the pipe are both given by the instantaneous rate of change of the volume.
V = 1200 + 1020t – 17t2
V’ = 1020 – 34t
a) @ 5 min
V’ = 1020 - 34×5 = 850 L/min
b) @ 10 min
V’ = 1020 - 34×10 = 680 L/min
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Model No. 3
The tangent, AB, touches the curve f(x) = 12 + 4x – x2 at point A, and cuts the x-axis at point B. If point A has an x-coordinate of 3, find
i) The length AB and ii) The acute angle AB makes with
the x-axis
X
Y
-3 -2 -1 1 2 3 4 5 6 7 8 9 10 11 12
3
6
9
12
15
0
A
B
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f(x) = 12 + 4x – x2 f(3) = 12 + 4×3 – 32 = 15
f’(x) = 4 – 2xf’(3) = 4 – 2×3 = -2 = m
∴ y = mx + c 15 = -2×3 + c c = 21 y = -2x + 21
The tangent, AB, touches the curve f(x) = 12 + 4x – x2 at point A, and cuts the x-axis at point B. If point A has an x-coordinate of 3, find
i) The length AB and
ii) The acute angle AB makes with the x-axis
Find the equation of AB
find
(i) coordinates at A and
(ii) the gradient at A, then
(iii) equation of AB
X
Y
-3 -2 -1 1 2 3 4 5 6 7 8 9 10 11 12
3
6
9
12
15
0
A
B
(3, 15)
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y = -2x + 21
Cuts x-axis at y = 0∴ 0 = -2x + 21
x = 10.5
The tangent, AB, touches the curve f(x) = 12 + 4x – x2 at point A, and cuts the x-axis at point B. If point A has an x-coordinate of 3, find
i) The length AB and
ii) The acute angle AB makes with the x-axis
Now find where AB cuts the x-axis
X
Y
-3 -2 -1 1 2 3 4 5 6 7 8 9 10 11 12
3
6
9
12
15
0
A
B
(3, 15)
(10.5, 0)
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The tangent, AB, touches the curve f(x) = 12 + 4x – x2 at point A, and cuts the x-axis at point B. If point A has an x-coordinate of 3, find
i) The length AB and
ii) The acute angle AB makes with the x-axis
Now find the length of AB
X
Y
-3 -2 -1 1 2 3 4 5 6 7 8 9 10 11 12
3
6
9
12
15
0
A
B
(3, 15)
(10.5, 0)
(1DP) 8.16
)5.103()015(
)()(
22
212
212
xxyyd
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The tangent, AB, touches the curve f(x) = 12 + 4x – x2 at point A, and cuts the x-axis at point B. If point A has an x-coordinate of 3, find
i) The length AB and
ii) The acute angle AB makes with the x-axis
X
Y
-3 -2 -1 1 2 3 4 5 6 7 8 9 10 11 12
3
6
9
12
15
0
A
B
(3, 15)
(10.5, 0)θ
15 units
10.5 – 3 = 7.5 units
4.63
5.7
15tan
5.7
15tan
tan
1
adj
opp
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Exercise
Set 8.6 Page 277No. 1-3, 8, 11-13
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1. If y = x2 + Bx + 6 has a stationary point at (2,k) , find B and k.
2. Sketch a curve for which, when x< -2
when -2<x<1
when x>1 negativeis
positiveis
negativeis
dxdy
dxdy
dxdy
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3. For the first 4 seconds of its motion, a particle moving in a straight line has a velocity at time t seconds, given by
v = t3 - 5t2 + 6t + 1 m/s.
Find the greatest and least velocity of the particle in this time.
4. A man standing on a cliff above the ocean throws a ball directly upward. The height of the ball above the water t seconds after release is h metres where h = 50+70t-10t2.
i. How high above the water will the ball go?
ii. What is the time of flight?
iii. Comment upon any unusual results?
5. Find, with reasons, the greatest and least values of the function f(x) = x3 – 2x2 in the interval -1 x 1
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6.Show that the curves whose equations are given by
y = and y = have the same slope at their point of intersection.
Find the equation of the common tangent at this point.
7. Find the equation of the tangent to the curve y = x3 – 5 at the point (1,-4).
Where does this tangent cut the x-axis?
8. A manufacturer makes a batch of N articles, the cost of each one being N2 – 6N + 35 cents. If he sells each article for 50 cents, find an expression representing his profit, P cents, for the entire batch.
How many articles should be in the batch to maximise P?
Is it true that P is greatest when the cost per article is least?
13 x 25 xx
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9. Draw a neat sketch of the gradient function for each of the following.
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10. The concentration of adrenalin in the bloodstream immediately after stimulus is given by the function
where t is time in seconds and concentration is in ml/L. Find how long it takes for the concentration to reach its maximum and what that concentration is.
9
2.1)(
2 t
ttc