At a certain bakery, the price of each doughnut is $1.50. Let the random variable D represent the number of doughnuts a typical
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The given height of the cylinder of 1.5 m, and radius of 1 m, and the rate
of dripping of 110 cm³/s gives the following values.
1) The rate of change of the oil's radius when the radius is 0.5 m is r' ≈ <u>9.34 × 10⁻⁵ m/s</u>
2) The rate of change of the oil's height when the height is 20 cm is h' ≈ <u>1.97 × 10⁻³ m/s</u>
3) The rate the oil radius is changing when the radius is 10 cm is approximately <u>0.175 m/s</u>
<h3>How can the rate of change of the radius & height be found?</h3>The given parameters are;
Height of the tank, h = 1.5 m
Radius of the tank, r = 1 m
Rate at which the oil is dripping from the tank = 110 cm³/s = 0.00011 m³/s
From the shape of the tank, we have;
Which gives;
h = 1.5·r
Which gives;
When r = 0.5 m, we have;
The rate of change of the oil's radius when the radius is 0.5 m is r' ≈ <u>9.34 × 10⁻⁵ m/s</u>
2) When the height is 20 cm, we have;
h = 1.5·r
r = 20 cm ÷ 1.5 = cm =
m
Which gives;
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When the height is 20 cm = 0.2 m, we have;
The rate of change of the oil's height when the height is 20 cm is h' ≈ <u>1.97 × 10⁻³ m/s</u>
3) The volume of the slick, V = π·r²·h
Where;
h = The height of the slick = 0.1 cm = 0.001 m
Therefore;
V = 0.001·π·r²
When the radius is 10 cm = 0.1 m, we have;
The rate the oil radius is changing when the radius is 10 cm is approximately <u>0.175 m</u>
Learn more about the rules of differentiation here: