Selected Homework 0 Answers (Plus 6-10): BC Calculus Review Assignment

[Maple Plot]

f(x) = (x^3+1)^(1/2), Df(x) = 3/2*x^2/(x^3+1)^(1/2)


[Maple Plot]

f(x) = (x^2+x-1)/(x^2-1), Df(x) = -(x^2+1)/(x^2-1)^2


[Maple Plot]

f(x) = -x*tan(x), Df(x) = -tan(x)-x*(1+tan(x)^2)


[Maple Plot]

f(x) = (x+1)^x, Df(x) = (x+1)^x*(ln(x+1)+x/(x+1))


`Problem 2A.`

Int(2/3/x^(1/3),x) = h(x)+C

``

Int(2/3/x^(1/3),x) = x^(2/3)+C

`Check:  `*h(x) = x^(2/3)

Dh(x) = D(x^(2/3))

Dh(x) = 2/3/x^(1/3)


`Problem 2B.`

Int(cos(x)/sin(x)^(1/2),x) = h(x)+C

``

Int(cos(x)/sin(x)^(1/2),x) = 2*sin(x)^(1/2)+C

`Check:  `*h(x) = 2*sin(x)^(1/2)

Dh(x) = D(2*sin(x)^(1/2))

Dh(x) = cos(x)/sin(x)^(1/2)


`Problem 2C.`

Int(cot(x)^(1/2)*csc(x)^2,x) = h(x)+C

Int(cot(x)^(1/2)*csc(x)^2,x) = -2/3*(cos(x)/sin(x))^(1/2)*cos(x)/sin(x)+C

Int(cot(x)^(1/2)*csc(x)^2,x) = -2/3*cot(x)^(3/2)+C

`Check:  `*h(x) = -2/3*cot(x)^(3/2)

Dh(x) = D(-2/3*cot(x)^(3/2))

Dh(x) = cot(x)^(1/2)*(1+cot(x)^2)


`Problem 2D.`

Int(x*(2*x+1)^(1/2),x) = h(x)+C

``

Int(x*(2*x+1)^(1/2),x) = 1/15*(2*x+1)^(3/2)*(-1+3*x)+C

`Check:  `*h(x) = 1/15*(2*x+1)^(3/2)*(-1+3*x)

Dh(x) = D(1/15*(2*x+1)^(3/2)*(-1+3*x))

Dh(x) = x*(2*x+1)^(1/2)


[Maple Plot]

x^2+3*x*y+y^3 = 10

Dy = -1/3*(2*x+3*y)/(x+y^2)


[Maple Plot]

f(x) = (x^2-4)^(1/2), ` and `*`x '`(t) = 5

`So, `*`f '`(x) = x/(x^2-4)^(1/2), ` and `*x(t) = 5*t+C

y = f(x), x = x(t)

y(t) = f(x(t))

`By the Chain Rule`, `y '`(t) = `f '`(x(t))*`x '`(t)

`When `, x = 3, `y '` = 5*`f '`(3)

`y '` = 3*5^(1/2)


[Maple Plot]

`Area ` = 4*x*y

1/144*x^2+1/16*y^2 = 1

`y = `*f(x) = 1/3*(-x^2+144)^(1/2)

Area(x) = 4/3*x*(-x^2+144)^(1/2)

`Area '`(x) = -8/3*(x^2-72)/(-x^2+144)^(1/2)

`Area '`(x) = 0

-x^2+72 = 0

x = -6*2^(1/2), 6*2^(1/2)

`y = `*f(6*2^(1/2)) = 2*2^(1/2)

Area(6*2^(1/2)) = 96


[Maple Plot]

x^3-6*x*y+y^3 = 0

dy/dx = (x^2-2*y)/(2*x-y^2)

x^2-2*y = 0

y = 1/2*x^2

-2*x^3+1/8*x^6 = 0

1/8*x^3*(-16+x^3) = 0

x = 0., 2.519842100

y = 0., 3.174802104

`Points of horizontal tangency: `*[x, y] = [0., 0.], [2.519842100, 3.174802104]

dx/dy = (-2*x+y^2)/(-x^2+2*y)

`So, by symmetry, Points of vertical tangency: `*[x, y] = [0., 0.], [3.174802104, 2.519842100]


[Maple Plot]

y^2-x^3*(2-x) = 0

y = (-x^2+2*x)^(1/2)*x, -(-x^2+2*x)^(1/2)*x

dy/dx = -x^2*(-3+2*x)/y

-x^2*(-3+2*x) = 0

x = 0., 1.500000000

y = 1.299038106, -1.299038106

`Points of horizontal tangency: `*[x, y] = [1.500000000, 1.299038106], [1.500000000, -1.299038106]

dx/dy = -y/x^2/(-3+2*x)

`Points of vertical tangency: `*[x, y] = [2, 0]


[Maple Plot]

x^(2/3)+y^(2/3) = 4

dy/dx = -y^(1/3)/x^(1/3)

`This clearly yields cusps and not points of tangency.`


[Maple Plot]

2*(x^2+y^2)^2-25*x^2+25*y^2 = 0

dy/dx = -x*(4*x^2+4*y^2-25)/y/(4*x^2+4*y^2+25)

-x*(4*x^2+4*y^2-25) = 0

So, x = 0, ` or `, -4*x^2-4*y^2+25 = 0

y = 1/2*(-4*x^2+25)^(1/2), -1/2*(-4*x^2+25)^(1/2)

1875/8-50*x^2 = 0

x = -2.165063509, 2.165063509

y = 1.250000001, -1.250000001

`Points of horizontal tangency: `

[-2.165063509, 1.250000001], [-2.165063509, -1.250000001]

[2.165063509, 1.250000001], [2.165063509, -1.250000001]

dx/dy = -y*(4*x^2+4*y^2+25)/x/(4*x^2+4*y^2-25)

y = 0

2*x^4-25*x^2 = 0

x = -3.535533906, 3.535533906

`Points of vertical tangency: `*[-3.535533906, 0], [3.535533906, 0]


[Maple Plot]

x^2*y^2-(y+1)^2*(4-y^2) = 0

dy/dx = -x*y^2/(x^2*y-3*y+2*y^3-4+3*y^2)

-x*y^2 = 0

So, x = 0, ` or `, y^2 = 0

`If `*x = 0, ` then`*y = -1, -2, 2

`Points of horizontal tangency: `*[0, -2], [0, 2]

dx/dy = -(x^2*y-3*y+2*y^3-4+3*y^2)/x/y^2

x^2 = (3*y-2*y^3+4-3*y^2)/y

(3*y-2*y^3+4-3*y^2)*y-(y+1)^2*(4-y^2) = 0

y^4+y^3+4*y+4 = 0

(y+1)*(y^3+4) = 0

y = -1, -2^(2/3)*` = `, -1.587401052

x = ((3*y-2*y^3+4-3*y^2)/y)^(1/2), -((3*y-2*y^3+4-3*y^2)/y)^(1/2)

x = 0, 0, .4501964649, -.4501964649

`Points of vertical tangency: `*[.4501964649, -1.587401052], [-.4501964649, -1.587401052]