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# AP Calculus AB - Special Assignment

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#### Transcript of AP Calculus AB - Special Assignment

AP Calculus Special Assignment By Eddie Hobaica and Tommy Lee Related Rates Volume of an elliptical cone = (3.14[(2r)^2]h)/6

Pi = 3.14

Volume of the elliptical cone is increasing at the rate of 18 in^3/sec. When the diameter of the elliptical cone is 4in., its volume is 8 in^3 and the radius is increasing at 1/4 in/sec. (a) At the instant when the diameter of the elliptical cone is 4 inches, what is the height h?

(b) At the instant when the diameter of the elliptical cone is 4 inches, what is the rate of change of the area of the base?

(c) At the instant when the diameter of the elliptical cone is 4 inches, what is the rate of change of its height?

(d) At the instant when the diameter of the elliptical cone is 4 inches, what is the instantaneous rate of change of the area of its base with respect to its height? Moving Particle A particle is moving along the x-axis with the velocity given by the function v(t) = -5t^4 + 9t^2 + 20 when t is greater than or equal to 0s but less than or equal to 5s.

x(t) = 0m when t = 2s (a) Find the particular solution for the position formula x(t). (Hint: Find the missing constant c)

(b) What is the acceleration (m/s^2) of the particle at t =1s?

(c) When does the particle change direction?

(d) What is the position of the particle at t = 4s? Answers Area of base = 3.14r^2

dV/dt = 18(3.14)in^3/sec

r = 2in

V = 8(3.14)in^3

dr/dt = (1/4)in/sec

dA/dt = (3.14)(2r)(dr/dt) = (3.14)[2(2in)(1/4in/sec)] = 3.14in^2/sec h = 6V/(3.14)(2r)^2 = 6V/4(3.14)r^2

dh/dt = ([(3.14)(2r)^2 x 6dv/dt] - [6V x (4(3.14) x 2r(dr/dt)])/[3.14(2r)^2]^2

dh/dt = [1728(3.14) - 192(3.14)^2]/[256(3.14)^2]in/sec

dh/dt = [27-3(3.14)]/[4(3.14)]in/sec dA/dh = (dA/dt)/(dh/dt) = (3.14in^2/sec)/([27-3(3.14)]/4[3.14])in *Related Rates d = 4in or r = 2in

V = [(3.14)([2r]^2)(h)]/6

8in^3 = [(3.14)([2(2in)]^2)(h)]/6

48in^3/[(3.14)(16in^2)] = h

h = 3/3.14 Answers a(t) = -20t^3 + 18t

a(1) = -20(1)^3 + 18(1)

a(1) = -2m/s^2 Use calculator to find when v(t) equals zero and see if v(t) changes sign.

v(t) = 0 at t = 1.7587 sec

v(1) = -5(1)^4 + 9(1)^2 + 20

v(1) = -5 + 9 + 20

v(1) = 24m/s

v(t) is positive when t is greater than or equal to 0 but less than 1.7587.

v(2) = -5(2)^4 + 9(2)^2 + 20

v(2) = -80 + 36 + 20

v(2) = -24m/s

v(t) is negative when t is less than or equal to 5 but greater than 1.7587. x(4) = -(4)^5 + 3(4)^3 + 20(4) - 32

x(4) = -784m *Moving Particle v(t) = -5t^4 + 9t^2 + 20

x(t) = -t^2 + 3t^3 + 20t + c

x(t) = 0 when t = 2

0 = (-2)^5 + 3(2)^3 + 20(2) + c

0 = -32 + 24 + 40 + c

c = -32 Areas and Volume The two functions below cross to form the shaded area.

Blue function: g(x) = -(x-5)^2 - 2x + 30

Red function: f(x) - (x-5)^2 + 2x (a) Find the maximum and minimum points on the shaded area.

(b) Find the solutions (Hint: Points of intersection).

(c) Find the area enclosed using vertical sums (Units are m^2).

(d) Set-up and evaluate the integral that gives the volume of the solid formed by revolving the region about line y = 9 (Units are m^3). Answers *Areas and Volume f(x) = -(x-5)^2 - 2x + 30

g(x) = (x-5)^2 + 2x Area = Integral Set f(x) = g(x)

-(x-5)^2 - 2x + 30 = (x-5)^2 + 2x

-x^2 +10x - 25 - 2x + 30 = x^2 - 10x + 25 + 2x

2x^2 - 20x + 50 + 4x - 30 = 0

2x^2 - 16x + 20 = 0

2(x^2 - 8x + 10) = 0 Use quadratic equation to find the solutions.

x = 4 + squareroot{6}

x = 4 - squareroot{6} [-(x-5)^2 - 2x + 30] - [(x-5)^2 + 2x] 4 + squareroot{6}

4 - squareroot{6} Limits used in next problem Area = 39.19m^2 Volume = 3.14 Integral 4 + squareroot{6}

4 - squareroot{6} [f(x)]^2 - [g(x)]^2 dx |

| |

| Volume = 3.14 Integral 4 + squareroot{6}

4 - squareroot{6} |

| [-(x-5)^2 - 2x + 30]^2 - [(x-5)^2 +2x]^2 dx f(x)

g(x) Input the integral into your calculator to find the answer. Final Volume is about: 1175.7551 x (3.14) = 3691.871m^3 (a) (b) (c) (d) (a) (b) (c) (d) f(x) = -(x-5)^2 - 2x + 30

df(x)/dy = -2(x-5) - 2

0 = -2(x-5) - 2

-1 = (x-5)

x = 4 g(x) = (x-5)^2 + 2x

dg(x)/dy = 2(x-5) + 2

0 = 2(x-5) + 2

-1 = (x-5)

x = 4 f(4) = -[(4-5)^2] - 2(4) +30

f(4) = 21

Maximum: (4,21) g(4) = [(4-5)^2 + 2(4)]

g(4) = 9

Minimum: (4,9) (a) (b) (c) (d)

Full transcriptPi = 3.14

Volume of the elliptical cone is increasing at the rate of 18 in^3/sec. When the diameter of the elliptical cone is 4in., its volume is 8 in^3 and the radius is increasing at 1/4 in/sec. (a) At the instant when the diameter of the elliptical cone is 4 inches, what is the height h?

(b) At the instant when the diameter of the elliptical cone is 4 inches, what is the rate of change of the area of the base?

(c) At the instant when the diameter of the elliptical cone is 4 inches, what is the rate of change of its height?

(d) At the instant when the diameter of the elliptical cone is 4 inches, what is the instantaneous rate of change of the area of its base with respect to its height? Moving Particle A particle is moving along the x-axis with the velocity given by the function v(t) = -5t^4 + 9t^2 + 20 when t is greater than or equal to 0s but less than or equal to 5s.

x(t) = 0m when t = 2s (a) Find the particular solution for the position formula x(t). (Hint: Find the missing constant c)

(b) What is the acceleration (m/s^2) of the particle at t =1s?

(c) When does the particle change direction?

(d) What is the position of the particle at t = 4s? Answers Area of base = 3.14r^2

dV/dt = 18(3.14)in^3/sec

r = 2in

V = 8(3.14)in^3

dr/dt = (1/4)in/sec

dA/dt = (3.14)(2r)(dr/dt) = (3.14)[2(2in)(1/4in/sec)] = 3.14in^2/sec h = 6V/(3.14)(2r)^2 = 6V/4(3.14)r^2

dh/dt = ([(3.14)(2r)^2 x 6dv/dt] - [6V x (4(3.14) x 2r(dr/dt)])/[3.14(2r)^2]^2

dh/dt = [1728(3.14) - 192(3.14)^2]/[256(3.14)^2]in/sec

dh/dt = [27-3(3.14)]/[4(3.14)]in/sec dA/dh = (dA/dt)/(dh/dt) = (3.14in^2/sec)/([27-3(3.14)]/4[3.14])in *Related Rates d = 4in or r = 2in

V = [(3.14)([2r]^2)(h)]/6

8in^3 = [(3.14)([2(2in)]^2)(h)]/6

48in^3/[(3.14)(16in^2)] = h

h = 3/3.14 Answers a(t) = -20t^3 + 18t

a(1) = -20(1)^3 + 18(1)

a(1) = -2m/s^2 Use calculator to find when v(t) equals zero and see if v(t) changes sign.

v(t) = 0 at t = 1.7587 sec

v(1) = -5(1)^4 + 9(1)^2 + 20

v(1) = -5 + 9 + 20

v(1) = 24m/s

v(t) is positive when t is greater than or equal to 0 but less than 1.7587.

v(2) = -5(2)^4 + 9(2)^2 + 20

v(2) = -80 + 36 + 20

v(2) = -24m/s

v(t) is negative when t is less than or equal to 5 but greater than 1.7587. x(4) = -(4)^5 + 3(4)^3 + 20(4) - 32

x(4) = -784m *Moving Particle v(t) = -5t^4 + 9t^2 + 20

x(t) = -t^2 + 3t^3 + 20t + c

x(t) = 0 when t = 2

0 = (-2)^5 + 3(2)^3 + 20(2) + c

0 = -32 + 24 + 40 + c

c = -32 Areas and Volume The two functions below cross to form the shaded area.

Blue function: g(x) = -(x-5)^2 - 2x + 30

Red function: f(x) - (x-5)^2 + 2x (a) Find the maximum and minimum points on the shaded area.

(b) Find the solutions (Hint: Points of intersection).

(c) Find the area enclosed using vertical sums (Units are m^2).

(d) Set-up and evaluate the integral that gives the volume of the solid formed by revolving the region about line y = 9 (Units are m^3). Answers *Areas and Volume f(x) = -(x-5)^2 - 2x + 30

g(x) = (x-5)^2 + 2x Area = Integral Set f(x) = g(x)

-(x-5)^2 - 2x + 30 = (x-5)^2 + 2x

-x^2 +10x - 25 - 2x + 30 = x^2 - 10x + 25 + 2x

2x^2 - 20x + 50 + 4x - 30 = 0

2x^2 - 16x + 20 = 0

2(x^2 - 8x + 10) = 0 Use quadratic equation to find the solutions.

x = 4 + squareroot{6}

x = 4 - squareroot{6} [-(x-5)^2 - 2x + 30] - [(x-5)^2 + 2x] 4 + squareroot{6}

4 - squareroot{6} Limits used in next problem Area = 39.19m^2 Volume = 3.14 Integral 4 + squareroot{6}

4 - squareroot{6} [f(x)]^2 - [g(x)]^2 dx |

| |

| Volume = 3.14 Integral 4 + squareroot{6}

4 - squareroot{6} |

| [-(x-5)^2 - 2x + 30]^2 - [(x-5)^2 +2x]^2 dx f(x)

g(x) Input the integral into your calculator to find the answer. Final Volume is about: 1175.7551 x (3.14) = 3691.871m^3 (a) (b) (c) (d) (a) (b) (c) (d) f(x) = -(x-5)^2 - 2x + 30

df(x)/dy = -2(x-5) - 2

0 = -2(x-5) - 2

-1 = (x-5)

x = 4 g(x) = (x-5)^2 + 2x

dg(x)/dy = 2(x-5) + 2

0 = 2(x-5) + 2

-1 = (x-5)

x = 4 f(4) = -[(4-5)^2] - 2(4) +30

f(4) = 21

Maximum: (4,21) g(4) = [(4-5)^2 + 2(4)]

g(4) = 9

Minimum: (4,9) (a) (b) (c) (d)