Presentation on theme: "Mathematics. Session Applications of Derivatives - 1."— Presentation transcript:
Session Applications of Derivatives - 1
Session Objectives Rate of Change of Quantities Slope and Equation of Tangent Slope and Equation of Normal Angle Between Two Curves
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
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
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
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.
Solution Let x be the edge of the variable cube and V be the volume at any time t.
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.
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
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
Normal Equation of normal to the curve at (x 0, y 0 ) is As normal is perpendicular to tangent at the point of contact
Example-3 Find the equation of the tangent and normal to the curve y = x 4 – 6x x 2 – 10x + 5 at (0,5). Solution :
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
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.
Con. The tangent is parallel to Therefore, the curve becomes (1, –6) lies on (i)
Angle Between Two Curves
The other angle is (1) Orthogonal curves: m 1 m 2 = - 1 (2) Curves touch each other: m 1 = m 2
Example-5 Show that the curves x 2 = 4y and 4y + x 2 = 8 intersect orthogonally at (2, 1).
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