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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 = x 0 Let be a function of x. Rate 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. 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 = t 0 can be written as at t = t 0

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Rate of Change of Quantities If both x and y are functions of t, then Rate of change of y with respect to t 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 Let x be the edge of the variable cube and V be the volume at any time t.

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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. = 1.6 x 22 = 35.2 cm 2 /sec

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Tangent Let be a continuous curve and let (x 0, y 0 ) be a point on the curve. The slope of the tangent to curve f(x) at (x 0, y 0 ) is The equation of the tangent to the curve at (x 0, y 0 ) is

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Normal Equation of normal to the curve at (x 0, y 0 ) is As normal is perpendicular to tangent at the point of contact

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Example-3 Find the equation of the tangent and normal to the curve y = x 4 – 6x 3 + 13x 2 – 10x + 5 at (0,5). Solution :

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Solution Cont. Equation of tangent at (0, 5) is y – 5 = -10 (x – 0) Slope of the normal at (0, 5) Equation of normal at (0, 5) is

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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.

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Con. The tangent is parallel to Therefore, the curve becomes (1, –6) lies on (i)

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Angle Between Two Curves

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The other angle is 180 0 - (1) Orthogonal curves: m 1 m 2 = - 1 (2) Curves touch each other: m 1 = m 2

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Example-5 Show that the curves x 2 = 4y and 4y + x 2 = 8 intersect orthogonally at (2, 1).

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Solution We have x 2 = 4y and 4y + x 2 = 8 Hence, the curves intersect orthogonally at (2, 1). m 1 m 2 = 1 x (-1) = -1

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Thank you

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Implicit Differentiation 3.6. Implicit Differentiation So far, all the equations and functions we looked at were all stated explicitly in terms of one.

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