# 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?

### 4 Answers

- BrainardLv 73 years agoBest 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

- King LeoLv 73 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

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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

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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

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- cidyahLv 73 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

- Sky - VhinLv 63 years ago
you can search in google, there are many students who asked questions like this before .........