Calculus - Related Rates questions?

I have been practicing these problems but have no idea how to go about them. Can you please solve these showing all your work so I can study from it? Thanks in advance

1) If a snowball melts so that its surface area decreases at a rate of 5 cm2/min, find the rate at which the diameter decreases when the diameter is 9 cm. (Give your answer correct to 4 decimal places.)

2) A kite 100 ft above the ground moves horizontally at a speed of 11 ft/s. At what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out?

3) Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the equation PV = C, where C is a constant. Suppose that at a certain instant the volume is 600 cm3, the pressure is 200 kPa, and the pressure is increasing at a rate of 30 kPa/min. At what rate is the volume decreasing at this instant?

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  • 3 years ago
    Best Answer

    1) If a snowball melts so that its surface area decreases at a rate of 5 cm2/min, find the rate at which the diameter decreases when the diameter is 9 cm. (Give your answer correct to 4 decimal places.)

    Let x = diameter

    Then

    A = 4π(x/2)^2 ..............area of a sphere

    = 4πx^2/4

    = πx^2

    dA/dt = π(2x)dx/dt

    dx/dt = dA/dt/2πx

    = - 5/2π(9)

    = - 0.08841941288

    = - 0.0884 cm/min

    2) A kite 100 ft above the ground moves horizontally at a speed of 11 ft/s. At what rate is the angle (in radians) between the string and the horizontal decreasing when 200 ft of string have been let out?

    tan(θ) = 100/11t

    sec^2(θ)dθ/dt = -100/11t^2

    dθ/dt = - 100/11t^2 * cos^2(θ)............(1)

    sin(θ) = 100/200

    = 1/2

    θ = π/6...............(2)

    200^2 - 100^2 = (11t)^2

    30000 = 121t^2

    t^2 = 247.9339...............(3)

    Substitute t^2 and θ in (1)

    dθ/dt = - 0.0275 rad/s

    3) Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and volume V satisfy the equation PV = C, where C is a constant. Suppose that at a certain instant the volume is 600 cm3, the pressure is 200 kPa, and the pressure is increasing at a rate of 30 kPa/min. At what rate is the volume decreasing at this instant?

    PV = C

    VdP/dt + PdV/dt = 0

    PdV/dt = - VdP/dt

    dV/dt = - V/P * dP/dt

    = - 600/200 * 30

    = - 90 cm^3/min

  • 3 years ago

    .

    1.

    Surface area of a sphere:

    A = 4πr²

    but Diameter = 2 * radius

    r = ½D

    A = 4π(½D)²

    A = πD²

    ➤find the rate at which the diameter decreases … means find dD/dt where D is the diameter at a particular time

    A = πD²

    dA/dt = 2πD dD/dt

    -5 cm²/min = 2π(9 cm) dD/dt

    dD/dt = -0.0884 cm/min

    ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

    2.

    dx/dt = 11 ft/s

    When 200ft of string is let out

    x² = 200² - 100² = 30000

    x = 100√3

    cosθ = (100√3)/200

    cos²θ = ¾

    tanθ = 100/x

    sec²θ dθ/dt = -100/x² dx/dt

    dθ/dt = -(100/x²) (cos²θ) dx/dt

    dθ/dt = -(100/30000) (¾) (11)

    dθ/dt = -0.0275 rad/s

    decreasing at a rate of 0.0275 rad/s

    ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

    https://www.flickr.com/photos/32696818@N08/2994819...

    3.

    PV = C

    P = 200kPa, V = 600cm³ and dP/dt = 30 kPa/min

    PV = C

    PdV/dt + VdP/dt = 0

    dV/dt = - (V/P) dP/dt

    dV/dt = - (600cm³ / 200kPa) (30 kPa/min)

    dV/dt = -90cm³/min

    volume is decreasing at a rate of 90cm³/min

    ⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯

  • cidyah
    Lv 7
    3 years ago

    1)

    Surface area of a sphere = 4pi r^2

    2r=d

    r = d/2

    Let S be the surface area

    S = 4pi (d/2)^2

    S = pi d^2

    dS/dt = 2 pi d (dd/dt)

    dS/dt = -5 cm^2/min (given)

    when d=9

    -5 = 2 pi (9) (dd/dt)

    dd/dt = -5 /(18pi) cm/min

    2)

    y=100

    z=200

    x^2+y^2=z^2

    x= sqrt(200^2-100^2) = 173.2051

    sin(θ) = x/z =173.2051/200

    θ = arcsin(173.2051/200) = 1.0472 radians

    sin(θ) = x/z

    z sin(θ) = x

    dz/dt sin(θ) + z cos(θ) dθ/dt = dx/dt

    (0) + (200) cos(1.0472) dθ/dt = 11

    dθ/dt = 11 / (200 cos(1.0472)) = 0.11 radians/s

    https://gyazo.com/ee533356f759549529ed24a57cde83b7

    3)

    PV = C

    P dV/dt + V dP/dt = 0

    (200) dV/dt + (600) (30) = 0

    solve for dV/dt = -90 m^3 /min

  • 3 years ago

    you can search in google, there are many students who asked questions like this before .........

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