Calculus-II Lec Notes (Set 1) PDF

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SMA 2102: CALCULUS II⃝cFrancis O. Ochiengfrancokech@gmailDepartment of Pure and Applied MathematicsJomo Kenyatta University of Agriculture and TechnologyCourse Description Parametric and implicit differentiation involving higher order derivatives. Applications to equations of tangent line and normal...

Description

SMA 2102: CALCULUS II c Francis O. Ochieng [email protected]

Department of Pure and Applied Mathematics Jomo Kenyatta University of Agriculture and Technology

Course Description • Parametric and implicit differentiation involving higher order derivatives. equations of tangent line and normal line.

Applications to

• Curve sketching and asymptotes. • Hyperbolic functions: their definitions, their differentiation and integration. • Techniques of integration: powers of trigonometric functions, standard substitutions including trigonometric and hyperbolic functions, integration by parts, integration via partial fractions, integration via t-substitution. • Solutions of first order ordinary differential equations by separation of variables method. • Applications of integration to: kinematics including simple harmonic motion, economics, arc length, plane and surface area, and volume in Cartesian coordinates. • Numerical integration: mid-ordinate rule, trapezoidal rule and Simpson’s rule. • Complex numbers: Argand diagrams, arithmetic operators and geometric representation. De Moivre’s theorem and application to trigonometric identities, roots of complex numbers.

References [1] Calculus: Early Transcendentals (8th Edition) by James Stewart [2] Calculus with Analytic Geometry by Roland E. Larson, Robert P. Hostetler and Bruce H. Edwards; 5th edition [3] Calculus and Analytical Geometry (9th edition) by George B. Thomas and Ross L. Finney [4] Advanced Engineering Mathematics (10th ed.) by Erwin Kreyszig [5] Calculus by Larson Hostellem

Lecture 1

1 1.1

Parametric and Implicit Differentiation Parametric differentiation

Consider the curve defined by the functions x = x(t), y = y(t) 1

(1)

1.1

Francis Oketch c

Parametric differentiation

If the functions in equation (1) have the same domain, then the set of points (x, y) for t in this domain dy (or is called the parametric representation of the curve given by equation (1). The first derivative, dx ′ y ) is given by dy dy/dt = dx dx/dt d2 y dy is found in terms of the parameter t and Since (or y′′ ) requires differentiation with respect to dx dx2 x, we use the chain rule of differentiation. Thus, d dy ( ) ( ) d2 y d dy d dy dt dt dx = = = × dx dx/dt dx2 dx dx dt dx (

Similarly,

)

d3 y (or y′′′ ) is given by dx3 (

)

d d2 y ( ) ( ) dt dx2 d3 y d d2 y dt d d2 y = , = × = 2 2 3 dx dx dt dx dx dx/dt dx etc. Example(s): 1. If x = t3 + 1, y = t2 − t + 4, find

dy d2 y d3 y in terms of t. , and dx3 dx dx2

Solution dy dx = 3t2 , = 2t − 1. Therefore, chain rule yields dt dt dy dx d2 y dx2 d3 y dx3

dy/dt 2t − 1 = 3t2 dx/dt ( ) ( ) ( ) ( ) d dy dt 1 d 2t − 1 2(1 − t) 1 3t2 (2) − (2t − 1)(6t) = · · = = · = dt dx 9t4 dx dt (3t2 ) 3t2 (3t2 ) 9t5 =

(

)

d d2 y d 2 − 2t dt = = · 2 dx dt 9t5 dt dx 8t − 10 = 27t8 (

) (

·

(

1 9t5 (−2) − (2 − 2t)(45t4 ) = 2 3t 81t10 )

2. The parametric equations of a curve are x = et , y = sin t. Find Hence, show that x2

dy d2 y + x + y = 0. 2 dx dx

) (

·

1 3t2

)

d2 y dy and as functions of t. dx dx2

Solution dy dx = et , = cos t. Therefore, chain rule yields dt dt dy dx d2 y dx2 Now, x2

= =

dy/dt cos t = t = e−t cos t e dx/dt ( ) ) d dy dx d ( −t −e−t sin t − e−t cos t = e−2t (− sin t − cos t) ÷ = e cos t ÷ et = et dt dx dt dt

[ ] [ ] dy d2 y 2t −2t t −t + x + y = e e (− sin t − cos t) + e e cos t + sin t = − sin t − cos t + cos t + sin t dx2 dx = 0

2

1.1

Francis Oketch c

Parametric differentiation

3. Given the curve x = sin 2θ, y = cos 2θ. Find

dy

d2 y d3 y3 as functions of θ. dx and dx dx2 ,

Solution dy dx = 2 cos 2θ, = −2 sin 2θ. Therefore, chain rule yields dθ dθ dy dy/dθ −2 sin 2θ = − tan 2θ = = 2 cos 2θ dx dx/dθ ( ) ) ) ( d dy d dθ ( 1 d2 y 2 = = (− tan 2θ) · = −2 sec 2θ · = − sec3 2θ dx2 dx dx dθ dx 2 cos 2θ ( ) ) ) dθ ( ) ( d d2 y 1 d3 y d ( 3 3 = − sec 2θ · = −6 sec 2θ tan 2θ · = = −3 sec4 2θ tan 2θ dx3 dθ dx dx dx2 2 cos 2θ √ t dy = √ 1 − t2 ) for some constant A, show that . dx 1 + 1 − t2 d2 y 1 Hence, find the value of at t = √ . dx2 2

4. If x = A(t + sin −1 t), y = A(1 −

Solution Recall that if p = sin−1 t, then sin p = t. Differentiating with respect to t on both sides yields dp 1 dp 1 1 cos p = √ =1 ⇒ = =√ . dt dt cos p 1 − sin2 p 1 − t2 Now,

) ( √ ) ( dx 1 + 1 − t2 1 √ =A = A 1+ √ dt 1 − t2 1 − t2 ( ) dy 1 At = A 0 − (1 − t2 )−1/2 (−2t) = √ dt 2 1 − t2 √ t dy/dt At dy 1 − t2 ) = = = √ √ ∴ · ( √ 2 2 dx dx/dt 1−t A 1+ 1−t 1 + 1 − t2

Then, d2 y dx2

dt d t d dy d t √ √ = = = dt 1 + 1 − t2 dx dx dx dx 1 + 1 − t2 √ t √ (1 + 1 − t2 )(1) − (1 − t2 )−1/2 (−2t) 1 − t2 2 ) √ · ( = √ 2 2 (1 + 1 − t ) A 1 + 1 − t2 (

)

)

(

)

(

]

[

√ √ t2 √ 1 − t2 (1 + 1 − t2 ) + √ 1 − t2 1 1 − t2 + 1 − t2 + t2 = √ √ √ = = 2 3 2 3 A(1 + 1 − t2 )2 A(1 + 1 − t ) A(1 + 1 − t )

1 Therefore, at t = √ , we have 2 d2 y   dx2 t= √12

=

1 (

A 1+

=

=

√

1−(

) 1 2 2 √ ) 2

=

1 (

A 1+

√

1 1− 2

)2 =

1 (

A 1+

√

1 2

)2

(√ )2 2 2 − 1 2 1 ( )2 = ) = (√ )2 )2 (√ (√ )2 = (√ 1 2 2 + 1 2−1 A 2+1 2 + 1 A A 1+ √ √ A 2 2 )2 (√ 2 2−1 √ 2 = (3 − 2 2) 2 A A(2 − 1)

1

3

1.1

Francis Oketch c

Parametric differentiation

5. If x = A[cos t + ln(tan(t/2))] and y = A sin t for some constant A, find

d2 y dx2

π. at t = 4

Solution dx dt

1 1 cos(t/2) = A − sin t + · sec2 (t/2) · = A − sin t + 2 sin(t/2) cos2 (t/2) tan(t/2) 2 [

]

[

]

[

1 − sin2 t 1 A cos2 t =A = A − sin t + = sin t sin t sin t [

dy dt dy ∴ dx

]

]

= A cos t =

sin t dy/dt a cos t = tan t = A cos t · = 2 A cos2 t dx/dt A cos t sin t

Then, d2 y dx2

Therefore, at t =

d dy d = = (tan t) = dx dx dx sin t 1 = sec2 t · = · 2 A cos t cos2 t (

)

d dt (tan t) dt dx sin t sin t = 2 A cos t A cos4 t

π , we have 4

d2 y   dx2 t= π4

=

√ √ √ √ sin(π/4) 4 4 2 1/ 2 1/ 2 2 2 √ = √ = = = = A A/4 A cos4 (π/4) 2A A 2 A(1/ 2)4

2t −1 2t ( 1+t2 ), show that 6. If x = tan −1 ( 1−t 2 ) and y = sin

dy = 1. dx

Solution 2t . Differentiating 1 − t2 2 dx (1 − t )(2) − (2t)(−2t) dx . Making = the subject with respect to t on both sides yields sec2 x 2 2 (1 − t ) dt dt yields 2t Given, x = tan−1 ( 1−t 2 ). Taking tangent on both sides yields tan x =

2 − 2t2 + 4t2 dx (1 − t2 )(2) − (2t)(−2t) 1 1 · · = = = 2 1 + tan x sec2 x (1 − t2 )2 (1 − t2 )2 dt

1 2

1+

2(1 + t2 ) 2(1 + t2 ) 2(1 + t2 ) (1 − t2 )2 · = = 2 2 2 2 2 2 2 2 (1 − t ) + 4t (1 − t ) (1 − t ) + 4t (1 + t2 )2 2 = 1 + t2

4t (1 − t2 )2

·

2 − 2t2 + 4t2 (1 − t2 )2

=

2t . Differentiating with 1 + t2 2 (1 + t )(2) − (2t)(2t) dy dy = . Making the subject yields respect to t on both sides yields cos y 2 2 dt (1 + t ) dt

Given y = sin−1 (

dy = dt

2t 1+t2 ).

Taking sine on both sides yields sin y =

2 + 2t2 − 4t2 (1 + t2 )(2) − (2t)(2t) 1 1 · · =√ = √ 2 2 2 cos y (1 + t ) (1 + t2 )2 1 − sin y

2(1 − t2 ) 2(1 − t2 ) (1 + t2 ) · √ = (1 + t2 )(1 − t2 ) (1 + t2 )2 − 4t2 (1 + t2 ) (1 − t2 )2 2 = 1 + t2

= √

4

1 1−

4t2 (1 + t2 )2

·

2 + 2t2 − 4t2 (1 + t2 )2

1.1

Francis Oketch c

Parametric differentiation Therefore, chain rule yields dy

7. Differentiate tan−1 Solution

(√

1

dx + x2 x

(√

−1

)

=

dy/dt dx/dt

2/(1 + t2 ) = 1. = 2/(1 + t2 )

with respect to tan−1 x.

)

du 1 + x2 − 1 . Now, and v = tan−1 x. We need to find Let u = tan−1 x dv ) (√ √ 1 + x2 − 1 1 + x2 − 1 ⇒ tan u = . Differentiating with respect to Given u = tan−1 x x x on both sides yields ) ( √ x √ − ( 1 + x2 − 1)(1) x du 1 + x2 = . sec2 u dx x2 du the subject yields Making dx ) ( √ √ x2 x √ √ − ( 1 + x2 − 1) − ( 1 + x2 − 1)(1) x 2 1 1 du 1 + x2 · = = · 1+x 2 2 2 dx x sec u 1 + tan u x2 2 √ x √ √ − ( 1 + x2 − 1) 1 + x2 − 1 x2 1 1 + x2 √ · = √ √ · = x2 x2 + ( 1 + x2 − 1)2 x2 1 + x2 ( 1 + x2 − 1)2 1+ 2 √x √ 1 + x2 − 1 1 + x2 − 1 √ √ = √ √ = [x2 + ( 1 + x2 − 1)2 ] 1 + x2 [2 + 2x2 − 2 1 + x2 )] 1 + x2 √ √ 1 + x2 − 1 1 + x2 − 1 √ = √ = 2 2 2 2[(1 + x ) 1 + x − (1 + x )] 2(1 + x2 )( 1 + x2 − 1) 1 = 2(1 + x2 ) Also, given v = tan−1 x ⇒ tan v = x. Differentiating with respect to x on both sides yields dv dv = 1. Making the subject yields sec2 v dx dx 1 1 1 dv = = = 2 2 dx sec v 1 + tan v 1 + x2 Therefore, chain rule yields 1 du/dx 1 (1 + x2 ) du 1 2(1 + x2 ) = = · = = 2 1 dv 1 2(1 + x ) dv/dx 2 (1 + x2 ) Exercise: 1 1 = 2 sec θ and y+ = 2cosec θ. x y ( ) dy dy 1 + y2 dx and in terms of θ, and hence show that =− Find . dθ dθ dx 1 + x2

1. Given that x = sec θ+tan θ and y = cosec θ+cot θ, show that x+

2. If x =

1 + 2t dy 1+t and y = , find the value of at t = 0. 1 − 2t 1−t dx

3. If x = t − sin t and y = 1 − sin t, find

d2 y . dx2

5

1.2

Francis Oketch c

Implicit Differentiation

Lecture 2

1.2

Implicit Differentiation

An equation f(x, y) = c, on certain restricted ranges of the variables x and y, is said to be defined implicitly for some constant c. For instance, the equation xy + x − 2y = 1 is implicit since y is not dy the subject. To find (or y′ ), follow these steps: dx i) Differentiate x normally ii) Apply direct chain rule in differentiating y iii) Collect like terms and make

dy the subject dx

Example(s): 1. Find y′ given that xy + x − 2y = 1. Solution Differentiating the given equation with respect to x, we get xy ′ + y + 1 − 2y′ = 0 2. Find y′ when x =

⇒

y′ =

y+1 2−x

√ 5 given that 4x2 + 9y2 = 36.

Solution Differentiating the given equation with respect to x, we get 8x + 18yy ′ = 0

⇒

y′ =

√ √ 5 we have 4( 5)2 + 9y2 − 36 = 0 ⇒ 4 Therefore, y = ± . 3 √ √ √ − 5 −4 5 ′ = .  At the point ( 5, 4/3), y = 9(4/3) 3 √ √ √ −4 5 5 .  At the point ( 5, −4/3), y′ = = 3 9(−4/3)

When x =

−4x −8x = . 18y 9y 20 + 9y2 − 36 = 0

⇒

y2 =

16 . 9

Exercise: 1. Find y′ given x2 y − xy 2 + x2 + y2 = 0. 1.2.1

[ans: y′ =

y2 − 2xy − 2x ] x2 − 2xy + 2y

Higher order derivatives of implicit functions

Higher order derivatives of implicit functions may be obtained in either of the two ways.  Method I: differentiate implicitly the derivatives of one lower order and replace y′ by the relation previously found.  Method II: differentiate implicitly both sides of the given equation as many times as is necessary to produce the required derivative and eliminate all derivatives of lower order. This procedure is recommended only when a derivative of higher order at a given point is required.

6

1.2

Francis Oketch c

Implicit Differentiation

Example(s): 1. Find y′ and y′′ given that x2 − xy + y2 = 3. Solution Differentiating the given equation with respect to x, we get 2x − xy ′ − y + 2yy′ = 0

y′ =

⇒

2x − y x − 2y

Differentiating further with respect to x yields y′′ =

d 2x − y (x − 2y)(2 − y′ ) − (2x − y)(1 − 2y′ ) = dx x − 2y (x − 2y)2 (

)

2x − y − 3y ′ ′ ′ ′ ′ 2x − xy − 4y + 2yy − 2x + 4xy + y − 2yy 3xy − 3y x − 2y = = (x − 2y)2 (x − 2y)2 (x − 2y)2 6x2 − 3xy − 3xy + 6y2 6x2 − 6xy + 6y2 3x [2x − y] − 3y[x − 2y] = = (x − 2 y )3 (x − 2y)3 (x − 2y)3 2 2 6(x − xy + y ) 6(3) 18 = = (x − 2y)3 (x − 2y)3 (x − 2y)3 3x

= = =

∴ y′′ =

[

]

18 (x − 2y)3

2. Find the value of y′′ at the point (-1,1) on the curve x2 y + 3y − 4 = 0. Solution Differentiating the given equation with respect to x, we get 2xy + x2 y′ + 3y′ = 0. Putting x = −1, y = 1 yields y′ =

1 2

Differentiating further with respect to x yields 2xy ′ + 2y + x2 y′′ + 2xy ′ + 3y′′ = 0 Putting x = −1, y = 1 and y′ =

⇒

4xy ′ + 2y + x2 y′′ + 3y′′ = 0

1 yields y′′ = 0. 2

3. Find y′ and y′′ at x = 1 given x3 y + xy 3 = 2.

[ans: y′ = −1, y ′′ = 0]

Solution Putting x = 1 in the given equation we get y + y3 = 2

⇒

y3 + y − 2 = 0

⇒

(y − 1)(y2 + y + 2) = 0

⇒

y=1

So, we need to solve for y′ and y′′ at the point (1, 1). Differentiating the given equation with respect to x, we get x3 y′ + 3x2 y + y3 + 3xy 2 y′ = 0. Putting x = 1, y = 1 yields y′ = −1 Differentiating further with respect to x yields x3 y′′ + 3x2 y′ + 3x2 y′ + 6xy + 3y2 y′ + 3xy 2 y′′ + 3y2 y′ + 6xy (y′ )2 = 0 Putting x = 1, y = 1 and y′ = −1 yields y′′ = 0. Exercise: 7

1.3

Francis Oketch c

Applications of differentiation

1. Find y′ and y′′ given (a) x + xy + y = 2. (b) x3 − 3xy + y3 = 1

(c) 2e−x + ey = 3ex−y

(d) x2/3 + y2/3 = 5 (e) xy 2 + x2 y = 2 √ (f) xy = x − 2y

x2 − 9 x2 + 9 2 (h) x + 2xy + 3y2 = 1 (g) y2 =

2. Find y′ , y ′′ and y′′′ at (a) the point (2,1) on x2 − y2 − x = 1.

(b) the point (1,1) on x3 + 3x2 y − 6xy 2 + 2y3 = 0.

1.3 1.3.1

Applications of differentiation Equations of tangent and normal lines

Let P (x1 , y1 ) be a point on the curve y = f(x), as shown below. The equation of a tangent line to the curve at point P is given by y − y1 =m x − x1 where m =

⇒

y = m(x − x1 ) + y1 ,

dy   is the gradient of the tangent line. dx (x1 ,y1 )

A normal line to the curve y = f(x) is a line through P which is perpendicular to the tangent line at point P. If the gradient of the tangent line at P is m, then the gradient of a normal to a curve 1 at P is − since the product of their gradients m is −1. Thus, the equation of the normal line at P is given by y=−

1 (x − x1 ) + y1 m

Example(s): 1. Find the equation of the tangent line and normal line to the curve y2 = 4x at point (4,4). Solution Given the curve y2 = 4x. Clearly, the point (4,4) lies on the curve. Differentiating the given dy 4 2 curve with respect to x yields 2ydy = 4dx ⇒ = = . The gradient of the tangent y dx 2y 2 1 dy   = = . Hence, the equation of the line to the curve at point (4, 4) is given by m = dx (4,4) 4 2 tangent line at point (4, 4) is 1 y = (x − 4) + 4 2 8

⇒

y=

1 x+2 2

1.3

Francis Oketch c

Applications of differentiation

1 = −2. Hence, the The gradient of the normal line to the curve at point (4, 4) is given by − m equation of the normal line at point (4, 4) is y = −2(x − 4) + 4

⇒

y = −2x + 12

2. Find the equation of the normal line to the curve xy = 4 at x = 3. Solution 4 . Thus, the point of contact is (3,4/3). 3 dy y Differentiating the given curve with respect to x yields xdy + ydx = 0 ⇒ = − . The x dx dy  4/3 4 gradient of the tangent line to the curve at point (3, 4/3) is given by m =  =− =− . 9 dx (4,4) 3 9 The gradient of the normal line at point (3, 4/3) is . Hence, the equation of the normal line at 4 point (3, 4/3) is 4 9 ⇒ 12y − 27x + 65 = 0 y = (x − 3) + 4 3 Substituting x = 3 into the given curve xy = 4 yields y =

Exercise: 1. Find the equation of the tangent line to the ellipse

x2 y2 = 1 at the point (x1 , y1 ). + a2 b2

Solution y2 2xdx 2ydy x2 + 2 =0 Given 2 + 2 = 1. Differentiating yields b a a2 b

b2 x dy = − 2 . The gradient dx a y b 2 x1 dy   = − 2 . Hence, the of the tangent line to the curve at point (x1 , y1 ) is given by m = dx (x1 ,y1 ) a y1 equation of the tangent line at point (x1 , y1 ) is

y=−

b 2 x1 (x−x1 )+ y1 a2 y1

⇒

a2 y1 (y−y1 )+ b2 x1 (x−x1 ) = 0

⇒

⇒

a2 yy1 +b2 xx1 = b2 x21 +a2 y12

Dividing through by a2 b2 yields yy1 x21 y21 xx1 + = + 2 =1 a2 b2 a2 b Therefore,

xx1 yy1 + 2 = 1. b a2

2. Find the equation of the tangent line and the normal line at a point for which t = 2, given that t2 3t ,y = the parametric equations of a curve are x = [ans: tangent: y = 83 x − 4, normal: 1+t 1+t ] y = − 38 x + 25 12 1.3.2

The intersection of two curves

The angle between two curves is the angle between the tangents to the two curves at the point of intersection. Let C1 and C2 be two plane curves intersecting each other at P and let P T1 and P T2 be tangents to C1 and C2 respectively at point P.

9

1.3

Francis Oketch c

Applications of differentiation

Angle T1 P T2 is the angle of intersection of the curves C1 and C2 . Suppose P T1 and P T2 intersect the x-axis (OX) at L1 and L2 respectively. If angle T1 L1 X = ψ1 and angle T2 L2 X = ψ2 . Then, θ = ψ2 − ψ1 . Now, the slope of the tangent to C1 at P is m1 = tan ψ1 and the slope of the tangent to C2 at P is m2 = tan ψ2 . Thus, m2 − m1 m2 − m1 tan ψ2 − tan ψ1 θ = tan−1 ⇒ = tan θ = tan(ψ2 − ψ1 ) = 1 + m1 m2 1 + m1 m2 1 + tan ψ1 tan ψ2 π If θ = ± , then C1 and C2 intersect orthogonally. Thus C1 and C2 are orthogonal if m1 m2 = −1 . 2 (

)

Example(s): 1. Find the angle of intersection of the following curves. (a) x2 + y2 = 8 and x2 = 2y at (2,2). Solution From x2 + y2 = 8. Differentiating yields 2xdx + 2ydy = 0 m1 =

⇒

x dy = − . So, dx y

dy  2  = − = −1 dx (2,2) 2

From x2 = 2y. Differentiating yields 2xdx = 2dy m2 =

dy = x. So, dx

⇒

dy   =2 dx (2,2)

Hence, if θ is the angle of intersection of the two curves then tan θ =

2+1 m2 − m1 = −3 = 1 + m1 m2 1 + (−1)(2)

⇒

θ = tan−1 (−3) = 108.4o

(b) y2 = 4ax and x2 = 4by at a point different from the origin (0,0), where a = b. Solution At the point of intersection, the two curves are equal. Solving the two curves simultaneously: y2 y4 = 4by ⇒ from y2 = 4ax we have x = . Substituting into x2 = 4by yields 16a 4a ) ( ( ) 2 y4 y3 y3 − 4by = 0 ⇒ y − 4b = 0. Hence, y = 0 or − 4b = 0 ⇒ y = 2 2 16a 16a 16a2 4a2/3 b1/3 . When y = 0, x = 0 and when y (= 4a2/3 b1/3 , x = 4a)1/3 b2/3 . So, the points of intersection of the two curves are (0, 0) and 4a1/3 b2/3 , 4a2/3 b1/3 . But (0,0) is unwanted. (

)

Therefore, the curves intersect at P 4a1/3 b2/3 , 4a2/3 b1/3 . From y2 = 4ax. Differentiating yields 2ydy = 4adx

⇒

dy 2a . So, = y dx

dy  a1/3 2a = 1/3  = 2/3 1/3 dx P 4a b 2b dy x From x2 =...