Question

A melting ice layer A spherical iron ball 8 in. in diameter is coated with a layer of ice of uniform thickness. If the ice melts at the rate of 10 in³/min, how fast is the thickness of the ice decreasing when it is 2 in. thick? How fast is the outer surface area of ice decreasing?

Accepted Answer

First, identify the variables: Let r be the radius of the iron ball, which is 4 inches (since the diameter is 8 inches). Let x be the thickness of the ice layer. The total radius of the ice-covered ball is R = r + x. The volume V of the ice is given by the formula for the volume of a sphere: V = $$ \frac{4}{3} \pi R^3 $$ - $$ \frac{4}{3} \pi r^3 . $$Differentiate this with respect to time t to find the rate of change of volume: $$ \frac{dV}{dt} = 4 \pi R^2 \frac{dR}{dt} . $$Since the ice melts at a rate of 10 in³/min, $$ \frac{dV}{dt} = -10 . $$Substitute this into the differentiated volume equation: $$ -10 = 4 \pi R^2 \frac{dR}{dt} . To $$find $$ \frac{dx}{dt} $$, note that $$ \frac{dR}{dt} = \frac{dx}{dt} $$ because r is constant. Solve for $$ \frac{dx}{dt} $$ using the equation $$ -10 = 4 \pi (r + x)^2 \frac{dx}{dt} $$ with x = 2 inches. For the rate of change of the surface area, use the surface area formula for a sphere: A = $$ 4 \pi R^2 . $$Differentiate with respect to time: $$ \frac{dA}{dt} = 8 \pi R \frac{dR}{dt} . $$Substitute the known values to find $$ \frac{dA}{dt} $$ when x = 2 inches.

A melting ice layer A spherical iron ball 8 in. in diameter is coated with a layer of ice of uniform thickness. If the ice melts at the rate of 10 in³/min, how fast is the thickness of the ice decreasing when it is 2 in. thick? How fast is the outer surface area of ice decreasing?

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