(22) session 22 carboxylic acids & carboxylic acid derivatives ii
Mathematics. Session Applications of Derivatives - 1.
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Transcript of Mathematics. Session Applications of Derivatives - 1.
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Mathematics
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Session
Applications of Derivatives - 1
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Session Objectives
Rate of Change of Quantities
Slope and Equation of Tangent
Slope and Equation of Normal
Angle Between Two Curves
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Rate of Change of Quantities
represents the rate
of change of y with respect to x at x = x0
0
0x=x
dyor f' x
dx
Let y = ƒ x be a function of x.
δx small change in x
δy corresponding small change in y
δx 0
δy dylim = =
δx dxRate of change of y with respect to x
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Rate of Change of Quantities
Similarly, rate of change of velocity with respect to time t,
represents acceleration.dvdt
Hence, velocity of a point body is defined as the rate of change of displacement with respect to time t.
Velocity at a time t = t0 can be written as at t = t0
dsdt
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Rate of Change of Quantities
If both x and y are functions of t, then
dy dy dx dx= × = ƒ x
dt dx dt dt
Rate of change of y with respect to t
dy=
dx
x rate of change of x with respect to t
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Example - 1
An edge of a variable cube is increasing at the rate of 5 cm/s. How fast is the volume of the cube increasing when the edge is 6 cm long.
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Solution
2dV=15x
dt2dV
=3x 5dt
2 3
x=6
dV=15 6 =540 cm /s
dt
Let x be the edge of the variable cube and V be the volume at any time t.
2dV dx=3x
dt dt
3 dxV = x and =5cm/s [Given]
dt
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Example - 2
The radius of a spherical soap bubble is increasing at the rate of 0.2 cm/sec. Find the rate of increase of its surface area,when the radius is 7 cm.
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Solution
Let r be the radius of a spherical soap bubble and S be the surface area at any time t.
2 drThen, S = 4 r and =0.2 cm/s
dt
dA dr=8 r
dt dt
dA=8 r 0.2
dt
x=7
dS 22=1.6× ×7
dt 7
= 1.6 x 22 = 35.2 cm2/sec
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Tangent
y = ƒ xLet be a continuous curve and let (x0 , y0) be a point on the curve.
The slope of the tangent to curve f(x) at (x0, y0) is
0 0
0x , y
dyƒ x or
dx
The equation of the tangent to the curve at (x0, y0) is
0 0
0 0x , y
dyy - y = x - x
dx
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Normal
Equation of normal to the curve at (x0 , y0) is
0 0
0 0
x , y
1y - y =- x - x
dydx
As normal is perpendicular to tangent at the point of contact
0 0x , y
1 1Slope of normal=- i.e. -
dySlope of tangentdx
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Example-3
Find the equation of the tangent and normal to the curve
y = x4 – 6x3 + 13x2 – 10x + 5 at (0,5). Solution :
4 3 2y = x - 6x +13x - 10x+5
3 2dy= 4x - 18x +26x - 10
dx
3 2
0, 5
dy= 4 0 - 18 0 +26 0 - 10=-10
dx
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Solution Cont.
Equation of tangent at (0, 5) is y – 5 = -10 (x – 0)
y - 5=-10x 10x+y - 5=0
Slope of the normal at (0, 5)
1 1=- =
-10 10
Equation of normal at (0, 5) is
1y - 5= x - 0 10y =x+50
10
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Example-4
If the tangent to the curve at (1, -6) is parallel to the line x – y + 5 = 0, find the values of a and b.
3y = x +ax +b
3Solution: Given curve is y = x +ax +b
2dy= 3x +a
dx
21, -6
dyThe slope of the tangent at 1,- 6 = = 3 1 +a
dx
= a+3
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Con.
The tangent is parallel to x y 5 0 y x 5
a+3 =1 a= -2
Therefore, the curve becomes 3y = x - 2x +b ... i
(1, –6) lies on (i)
6 1 2 b b 5
a 2, b 5
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Angle Between Two Curves
1 2θ =θ - θ
1 2tanθ = tan θ - θ
P (x , y )0 0y = g(x) y = f(x)
21
y
xO
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Angle Between Two Curves
1 2
1 2
tanθ - tanθ=
1+tanθ tanθ
1 0 1 2 0 2Where tanθ = f' x =m and tanθ =g' x =m
The other angle is 1800 -
(1) Orthogonal curves: m1m2 = - 1
(2) Curves touch each other: m1 = m2
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Example-5
Show that the curves x2 = 4y and 4y + x2 = 8 intersect orthogonally at (2, 1).
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Solution
We have x2 = 4y and 4y + x2 = 8
dydx
dy2x = 4 and 4 +2x = 0
dx
x dy xdyand = -
dx 2 dx 2=
dy2 2dym = = =1 and m = =- =-11 2dx 2 2dx2,1 2,1
Hence, the curves intersect orthogonally at (2, 1).
m1m2 = 1 x (-1) = -1
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Thank you