Tangents and normals Remember, the tangent to a curve at a point is a straight line that just touches the curve at that point. The normal to a curve at a point is a straight line that is perpendicular to the tangent at that point. We can use differentiation to find the equation of the tangent or the normal to a curve at a given point. For example: Find the equation of the tangent and the normal to the curve y = x2 – 5x + 8 at the point P(3, 2).

Tangents and normals y = x2 – 5x + 8 At the point P(3, 2) x = 3 so: 4 The gradient of the tangent at P is therefore 4. Using y – y1 = m(x – x1) give the equation of the tangent at the point P(3, 2): y – 2 = 4(x – 3) y – 2 = 4x – 12 y – 4x + 10 = 0

Tangents and normals The normal to the curve at the point P(3, 2) is perpendicular to the tangent at that point. The gradient of the tangent at P is 4 and so the gradient of the normal is Using y – y1 = m(x – x1) give the equation of the tangent at the point P(3, 2):

Task 1 Find the equation of the tangent to the curve y = x2 - 7x + 10 at the point (2, 0). Find the equation of the normal to the curve y = x2 – 5x at the point (6, 6). Find the equations of the normals to the curve y = x + x3 at the points (0, 0) and (1, 2), and find the co-ordinates of the point where these normals meet. For f(x) = 12 – 4x + 2x2, find an equation of the tangent and normal, at the point where x = -1 on the curve with equation y = f(x).

Rate of Change of Quantities Let be a function of x. Rate of change of y with respect to x represents the rate of change of y with respect to x at x = x0

Rate of Change of Quantities Velocity at a time t = t0 can be written as at t = t0 Hence, velocity of a point body is defined as the rate of change of displacement with respect to time t. Similarly, rate of change of velocity with respect to time t, represents acceleration.

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 cm2/sec

Tangent Let 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 The equation of the tangent to the curve at (x0, y0) is

Normal As normal is perpendicular to tangent at the point of contact Equation of normal to the curve at (x0 , y0) is

Example-3 Find the equation of the tangent and normal to the curve y = x4 – 6x3 + 13x2 – 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

Angle Between Two Curves The other angle is 1800 - q (1) Orthogonal curves: m1m2 = - 1 (2) Curves touch each other: m1 = m2

Example-5 Show that the curves x2 = 4y and 4y + x2 = 8 intersect orthogonally at (2, 1).

Solution We have x2 = 4y and 4y + x2 = 8 m1m2 = 1 x (-1) = -1 Hence, the curves intersect orthogonally at (2, 1).