This week Tangent lines to a circle Find the slope of the tangent line to the circle 2 2 2 Homework #5 due Thursday 11:30 (x − 2) + (y + 3) = (13) Read sections 3.5 and 3.6 at the point (7; 9) using the following techniques. Worksheet #5 on Tuesday tangent line is perpendicular to the radial line. solve for y and differentiate implicit differentiation Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124 Using the radial line Solving for y Solving for y we get that (y + 3)2 = 169 − (x − 2)2: The center of the circle is (2; −3). The slope of the line q between the center and (7; 9) is y = −3 ± 169 − (x − 2)2: 9 − (−3) 12 mradial = = : As (7; 9) is on the top half of the circle we have that 7 − 2 5 q 2 The slope of the tangent line is given by y = −3 + 169 − (x − 2) : −1 −1 −5 m = = = : Differentiating (and using the chain rule twice) tangent m 12=5 12 radial 1 y 0 = 0 + (169 − (x − 2)2)−1=2(−2(x − 2))(1) 2 Evaluated when x = 7 we get 1 −5 −5 y 0 = (169 − (7 − 2)2)−1=2(−2(7 − 2)) = p = : 2 144 12 Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124 Implicit Differentiation Implicit Differentiation Now we treat y as a function of x and differentiate both sides (remembering to use the chain rule!). To emphasize that y is a function of x we write y(x): Evaluating 2 2 (y(x) + 3) + (x − 2) = 169: −2(x − 2) y 0(x) = : 2(y(x) + 3) d d (y(x) + 3)2 + (x − 2)2 = (169) : at the point (7; 9) we get dx dx −2(7 − 2) −5 0 y 0(7) = = : 2(y(x) + 3)y (x) + 2(x − 2) = 0: 2(9 + 3) 12 Now we solve for y 0 in terms of x and y. −2(x − 2) y 0(x) = : 2(y(x) + 3) Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124 Implicit Differentiation dy Use implicit differentiation to find dx if 4y 2 + x2 = 4 p Find the tangent line at the point at (1; − 3=2). Treat y as a function of x. 4y 2 + x2 = 4 Differentiate both sides. (Use the chain rule!). d d 0 4y 2 + x2 = (4) Solve for y in terms of x and y. dx dx Plug in values of x and y to find y 0 dy 8y + 2x = 0 dx dy −2x = dx 8y dy −x = dx 4y Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124 Lookingp at the graph we see the ellipse and its tangent line at (1; − 3=2). p At the point (1; − 3=2) we get dy −2x −2(1) 1 = = p = p : dx 8y 8(− 3=2) 2 3 So the equation of the tangent line is p 1 (y + 3=2) = p (x − 1): 2 3 Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124 x3 + y 3 = 6xy Suppose that d d x3 + y 3 = 6xy: x3 + y 3 = (6xy) dx dx dy dy 3x2 + 3y 2 = 6 x + y(1) dy dx dx Find dx : dy Find the equation of the tangent line at (3; 3). 3y 2 − 6x = 6y − 3x2 dx At which points on the curve do we have a vertical tangent dy 6y − 3x2 line? = dx 3y 2 − 6x dy 2y − x2 = : dx y 2 − 2x Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124 At which points on the curve do we have a vertical tangent line? Find the equation of the tangent line at (3; 3). The slope of the tangent line is The slope of the tangent line is 2y − x2 2y − x2 y 2 − 2x : y 2 − 2x If the tangent line is vertical then the denominator must be 0. Thus y 2 − 2x = 0 or x = y 2=2. Plugging in x = 3 and y = 3 we get When we plug this into the equation 3 3 2(3) − (3)2 −3 x + y = 6xy m = = = −1: (3)2 − 2(3) 3 we get Using the point slope formula the tangent line is (y 2=2)3 + y 3 = 6(y 2=2)y y − 3 = −1(x − 3): y 6 + 8y 3 = 24y 3 y 6 − 16y 3 = 0 Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124 y 3(y 3 − 16) = 0 there are two tangent lines at (0; 0) and one of them is vertical. y 3(y 3 − 16) = 0: Thus either y 3 = 0 or y 3 = 16. If y 3 = 0 then y = 0 and x = 0. 3 4=3 1 2 5=3 If y = 16 then y = 2 , and x = 2 y = 2 : Thus the two candidates for a vertical tangent line occur at (0; 0) and (25=3; 24=3): At (25=3; 24=3) dy 2(24=3) − (25=3)2 27=3 − 210=3 = = dx (24=3)2 − 2(25=3) 0 dy 0 At (0; 0) we get dx = 0 so we need to check something else. Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124 there is a vertical tangent line x = 25=3. and the line y − 3 = −(x − 3) is tangent to the graph at (−3; 3) Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124 Find all points on the curve dy −y − 2xy 2 x2y 2 + xy = 2 = dx 2x2y + x where the slope of the tangent line is −1. dy Now we set dy = −1: First we will find dx . Then we will find all solutions to the dx simultaneous equations dy −y − 2xy 2 dy −1 = = x2y 2 + xy = 2 and = −1: dx 2x2y + x dx d d 2x2y + x = y + 2xy 2 x2y 2 + xy = (2) dx dx dy dy 2 + − − 2 = x2(2y ) + y 2(x2)0 + x + y(x)0 = 0 2x y x y 2xy 0 dx dx dy dy x2(2y ) + 2xy 2 + x + y = 0 (x − y)(1 + 2yx) = 0 dx dx dy (2x2y + x) = −y − 2xy 2 x = y or 2yx = −1 dx Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124 Case 1: x = y Case 2: xy = −1=2 x2y 2 + xy = 2 x2x2 + xx = 2 If xy = −1=2 then x4 + x2 − 2 = 0 x2y 2 + xy = (−1=2)2 + (−1=2) = −1=4 6= 2: (x2 − 1)(x2 + 2) = 0 This case yields no points. Thus there are two points on the curve where the slope of the x2 − 1 = 0 tangent line is −1, (−1; −1) and (1; 1). x = ±1 Thus we have two points (−1; −1) and (1; 1): Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124 Looking at the graph we that at (−1; −1) and (1; 1) the slope of Implicit Differentiation the tangent line is −1. Treat y as a function of x. Differentiate both sides. (Use the chain rule!). Solve for y 0 in terms of x and y. Plug in values of x and y to find y 0 Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124 Find the derivative of y 5 + 3x2y 2 + 5x4 = 93 Find the equation of the tangent line at the point (2; 1): Find the equation of the tangent line at the point (2; 1): dy −6y 2x − 20x3 −6(1)2(2) − 20(2)3 −172 d d = = = : y 5 + 3x2y 2 + 5x4 = (93) dx 5y 4 + 6x2y 5(1)4 + 6(2)2 29 dx dx The equation of the tangent line at the point (1; 2) is 4 dy 2 dy 2 3 5y + 3 x 2y + y 2x + 20x = 0 −172 dx dx y − 1 = (x − 2): 29 dy (5y 4 + 6x2y) = −6y 2x − 20x3 dx dy −6y 2x − 20x3 = dx 5y 4 + 6x2y Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124 At how many points does the curve 2y 3 + y 2 − y 5 = x4 − 2x3 + x2 have a horizontal tangent line. Find the x coordinates of those points. Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124 d d 2y 3 + y 2 − y 5 = x4 − 2x3 + x2 dx dx dy dy dy 6y 2 + 2y − 5y 4 = 4x3 − 6x2 + 2x dx dx dx dy 6y 2 + 2y − 5y 4 = 4x3 − 6x2 + 2x dx dy 4x3 − 6x2 + 2x = dx 6y 2 + 2y − 5y 4 dy 2x(2x2 − 3x + 1) = dx 6y 2 + 2y − 5y 4 dy 2x(2x − 1)(x − 1) = dx 6y 2 + 2y − 5y 4 The horizontal tangents and occur when x = 0, x = 1=2 and x = 1. Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124 Use implicit differentiation to find the derivative of y = sin−1(x): dy 1 Taking the sin of both sides we get = dx cos(y) sin(y) = x: Now we write cos(y) in terms of x. cos2(y) + sin2(y) = 1 Then d d (sin(y)) = (x): we have dx dx cos2(y) = 1 − sin2(y) = 1 − x2 dy cos(y) = 1: and p dx cos(y) = 1 − x2: dy 1 1 = = : Then −1 dy 1 1 dx cos(y) cos(sin (x)) = = p : dx cos(y) 1 − x2 Let’s find a nicer formula. Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124 Inverse Trig Functions Use implicit differentiation to find the derivative of y = tan−1(x): d 1 Taking the tan of both sides we get sin−1(x) = p : dx 1 − x2 tan(y) = x: d −1 cos−1(x) = p : dx 1 − x2 Then d d d −1 1 (tan(y)) = (x): tan (x) = : dx dx dx 1 + x2 dy d −1 −1 sec2(y) = 1: csc (x) = p : dx dx x x2 − 1 2 2 d 1 dy cos (y) cos (y) 1 1 sec−1(x) = p : = = = = : dx 2 dx 1 cos2(y) + sin2(y) 1 + tan2(y) 1 + x2 x x − 1 d −1 cot−1(x) = : dx 1 + x2 Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124 Find the derivative of 0 −1 p f (x) = x tan−1(4x): Find f (x) when f (x) = x sin ( x): p p f 0(x) = x(sin−1( x))0 + (x)0 sin−1( x) 0 −1 0 0 −1 p f (x) = x(tan (4x)) + (x) tan (4x) 1 1 −1=2 −1 = x p p ( x ) + sin ( x): 1 − ( x)2 2 1 = x (4) + tan−1(4x): 1 + (4x)2 Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124 Use implicit differentiation to find the derivative of y = ln(x): First we exponentiate both sides to get We also have that if y = loga(x) then ( ) ey = eln(x) = x: ay = aloga x = x: Taking derivatives we get Taking derivatives d y d (e ) = (x) : d y d dx dx (a ) = (x): dx dx dy ey = 1: dx dy ln(a)ay = 1: dy dx = 1=ey : dx dy = 1= ln(a)ay : dy dx = 1=x: dx dy = 1= ln(a)x: d dx (ln(x)) = 1=x: dx Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124 Find the derivative of g(x) = ln(8x6 + 3x2) Find the derivatives of p By the chain rule m(x) = ln cos(x)(4x − 3)7 x + 3 1 g0(x) = (8x6 + 3x2)0 8x6 + 3x2 It is best to use our rules of logarithms before taking derivatives.