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)) = √ = . 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 √ 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 Looking√ at the graph we see the ellipse and its tangent line at (1, − 3/2).

√ At the point (1, − 3/2) we get

dy −2x −2(1) 1 = = √ = √ . dx 8y 8(− 3/2) 2 3

So the equation of the tangent line is

√ 1 (y + 3/2) = √ (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)) = = √ . 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) = √ . dx 1 − x2 tan(y) = x. d   −1 cos−1(x) = √ . 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) = √ . dx dx x x2 − 1

2 2 d   1 dy cos (y) cos (y) 1 1 sec−1(x) = √ . = = = = . 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 √ f (x) = x tan−1(4x). Find f (x) when f (x) = x sin ( x). √ √ f 0(x) = x(sin−1( x))0 + (x)0 sin−1( x)

0 −1 0 0 −1 √ f (x) = x(tan (4x)) + (x) tan (4x) 1 1 −1/2 −1 = x 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  √  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. 1 1 m(x) = ln cos(x)
+ 7 ln(4x − 3) + ln x + 3)
= (48x5 + 6x). 2 8x6 + 3x2 1 1 1 1 For any function f (x) the chain rule tells us m0(x) = (− sin(x)) + 7 (4) + . cos(x) 4x − 3 2 (x + 3) d 1 ln(f (x)) = f 0(x). dx f (x)

Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124 Logarithmic Differentiation The last problem suggests a useful trick. Suppose √ g(x) = cos(x)(4x − 3)7 x + 3. so m(x) = ln(g(x)). Then by the chain rule Sometime it is necessary (or easier) to take natural logs before taking the derivative. Then we proceed with implicit 1 differentiation. m0(x) = g0(x) and g0(x) = g(x)m0(x). g(x) We use this when the function has a product or quotient of a large number of In this case terms or  √  g0(x) = g(x)m0(x) = cos(x)(4x − 3)7 x + 3 there is an x in both the base and exponent.

 1 1 1 1  (− sin(x)) + 7 (4) + . cos(x) 4x + 3 2 (x + 3)

Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124

Power Rule Differentiate √ x3/4 x2 + 1 y = . (3x + 2)5(2x + 1) Prove the power rule by logarithmic differentiation. If f (x) = xn then Taking natural log of both sides we get ln(f (x)) = ln(xn) = n ln(x). 3 1 ln(y) = ln(x) + ln(x2 + 1) − 5 ln(3x + 2) − ln(2x + 1). 4 2 Taking derivatives we get

d d 3 1  d d (ln(y)) = ln(x)+ ln(x2+1)−5 ln(3x+2)−ln(2x+1) . (ln(f (x))) = (n ln(x)) . dx dx 4 2 dx dx

1 0 3 1 2x 1 1 = ( / ) + − ( ) − ( ). 1 0 n y 1 x 2 5 3 2 f (x) = . y 4 2 x + 1 3x + 2 2x + 1 f (x) x √ 3 1 2x 1 1  x3/4 x2 + 1 n n y 0 = (1/x)+ −5 (3)− (2) . f 0(x) = f (x) = xn = nxn−1. 4 2 x2 + 1 3x + 2 2x + 1 (3x + 2)5(2x + 1) x x

Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124 Derivatives of functions with exponents Use logarithmic differentiation to find the derivative of y = xx .

ln(y) = x ln(x). If a and b are constants then

d  b d d a = 0. (ln(y)) = (x ln(x)) . dx dx dx d   1 dy 1 af (x) = ln(a)af (x)f 0(x). = x + ln(x)(1). dx y dx x d   dy f (x)b = bf (x)b−1f 0(x). = y(1 + ln(x)). dx dx

d  g(x) dy x f (x) use logarithmic differentiation = x (1 + ln(x)). dx dx

Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124

If √ 2 4 2.3 √ 2 (2x − 5) x + 3(x + 1) If y = (2 x)3x +2x find y 0. y = (x − 2)(x5 − 2)(3 − 7x)x4 √ 2 ln(y) = ln((2 x)3x +2x ). find y 0. √ ! (2x − 5) x2 + 3(x4 + 1)2.3 √ ln(y) = ln . ln(y) = (3x2 + 2x) ln(2 x). (x − 2)(x5 − 2)(3 − 7x)x4

1 dy 1 1 √ 1 2 4 = (3x2 + 2x) √ 2 x−1/2 + (6x + 2) ln(2 x). ln(y) = ln(2x − 5) + ln(x + 3) + 2.3 ln(x + 1) − ln(x − 2) y dx 2 x 2 2

dy  1 1 √  − ln(x5 − 2) − ln(3 − 7x) − 4 ln(x). = y (3x2 + 2x) √ 2 x−1/2 + (6x + 2) ln(2 x) . dx 2 x 2    d d 1 2 4 dy √ 2 1 1 √ (ln(y)) = ln(2x−5)+ ln(x +3)+2.3 ln(x +1)−ln(x−2) = (2 x)3x +2x (3x2 + 2x) √ 2 x−1/2 + (6x + 2) ln(2 x)) . dx dx 2 dx 2 x 2  − ln(x5 − 2) − ln(3 − 7x) − 4 ln(x) .

Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124 p y(x) = (5 + 2x2) x4 + 3(x4 + 1)x . Find y 0(x).   1 dy 1 1 1 1 1 2 p 4 4 x = 2 + 2x + 2.3 4x3 − (1) ln(y(x)) = ln (5 + 2x ) x + 3(x + 1) . y dx 2x − 5 2 x2 + 3 x4 + 1 x − 2

2 1 4 4 1 1 1 ln(y(x)) = ln(5 + 2x ) + ln(x + 3) + x ln(x + 1). − 5x4 − (−7) − 4 . 2 x5 − 2 3 − 7x x d d  1  √ ! (ln(y(x))) = ln(5 + 2x2) + ln(x4 + 3) + x ln(x4 + 1) . dy (2x − 5) x2 + 3(x4 + 1)2.3  1 1 1 dx dx 2 = 2 + 2x dx (x − 2)(x5 − 2)(3 − 7x)x4 2x − 5 2 x2 + 3 1 dy 1 1 1 1 = ( )+ ( 3)+ ( 3)+( ) ( 4+ ). 2 4x 4 4x x 4 4x 1 ln x 1 1 1 1 1 1 y(x) dx 5 + 2x 2 x + 3 x + 1 +2.3 4x3 − (1) − 5x4 − (−7) − 4 . x4 + 1 x − 2 x5 − 2 3 − 7x x dy p = (5 + 2x2) x4 + 3(x4 + 1)x dx  1 1 1 1  (4x) + + x (4x3) + (1) ln(x4 + 1) . 5 + 2x2 2 x4 + 3 x4 + 1

Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124

Related Rates

Let A(t) be the area of the spill. Let r(t) be the radius of the spill. An oil platform explodes and spills oil into the Gulf of Mexico. A(t) = π(r(t))2. The oil spill grows in a circular fashion. If the radius of the spill is growing at a rate of 25 kilometers per day when the radius is 220 kilometers, how fast is the area of the spill growing? A0(t) = 2πr(t)r 0(t).

Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124 Strategy

0 Let t0 be the time when r(t0) = 220. We are asked to find A (t0). We have that

0 r(t0) = 220 and r (t0) = 25. 1 Draw a picture. 2 Choose notation. Plugging into A0(t) = 2πr(t)r 0(t) 3 Write an equation. 4 Differentiate using the chain rule. we get A0(t) = 2π(220)(25) = 11000π. 5 Substitute into the equation and solve.

Professor Christopher Hoffman Math 124 Professor Christopher Hoffman Math 124