If a snowball melts so that its surface area decreases at a rate of 5 cm2/min, find the rate (in cm/min) at which the diameter decreases when the diameter is 11 cm. (Round your answer to three decimal places.
Question
If a snowball melts so that its surface area decreases at a rate of 5 cm2/min, find the rate (in cm/min) at which the diameter decreases when the diameter is 11 cm. (Round your answer to three decimal places.
Solution
This problem can be solved using related rates in calculus. Here are the steps:
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First, we need to understand the relationship between the surface area (A) and the diameter (d) of a sphere. The formula for the surface area of a sphere is A = 4πr², where r is the radius. But since we're given the diameter, we can express the radius as r = d/2. Substituting this into the surface area formula gives us A = πd².
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We're given that dA/dt = -5 cm²/min. This is the rate at which the surface area is decreasing.
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We want to find dd/dt, the rate at which the diameter is decreasing, when d = 11 cm.
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To find this, we can differentiate the surface area equation with respect to time (t) to get a relationship between dA/dt and dd/dt. This gives us dA/dt = 2πd * dd/dt.
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Now we can substitute the given values into this equation to solve for dd/dt: -5 = 2π(11) * dd/dt.
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Solving for dd/dt gives us dd/dt = -5 / (2π(11)) = -0.072 cm/min.
So, the diameter is decreasing at a rate of approximately 0.072 cm/min when the diameter is 11 cm.
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