Question

A spherical iron ball 12 in. in diameter is coated with a layer of ice of uniform thickness. If the ice melts at a rate of 11 in.^3/min how fast is the thickness of the ice decreasing when it is 3 in.​ thick? How fast is the outer surface area of ice​ decreasing?

a) Write an equation relating the volume of the sphere, V, to the thickness of the ice, x.

b) Differentiate both sides of the equation with respect to t, and solve for dx/dt.

c) The thickness of the ice is decreasing at a rate of ______________ when the ice is 3 in. thick.

d) Write an equation relating the outer surface area of the ice, A, to the thickness of the ice, x.

e) Differentiate both sides of the equation with respect to t.

f) The outer surface area of ice is decreasing at a rate of _______________ when the is 3 in. thick.

Answer 1

Refer to the step-by-step explanation. Follow along carefully, answers to each part are encased by two boxes.

Explanation:

Part (a): The volume of the sphere, "V," can be expressed as:

`\rightarrow V=(4)/(3) \pi r^3`

Where "r" is the radius of the sphere. Since the diameter of the iron ball is 12 inches, the radius is half of that, which is 6 inches. We can rewrite the equation in terms of the thickness of the ice, x, as follows:

`\rightarrow \boxed{\boxed{V = \frac43 \pi\big[(r + x)^3-(r)^3\big]}}`

Part (b): Let's differentiate both sides of the equation with respect to time, t:

`V = \frac43 \pi\big[(r + x)^3-(r)^3\big]\\\\\\\\\Longrightarrow V = \frac43 \pi\big[3r^2x+3rx^2+x^3]\\\\\\\\\Longrightarrow (dV)/(dt) = \frac43 \pi\big[3r^2+6rx+3x^2] \cdot (dx)/(dt)`

To find dx/dt, we need to solve for it. The left side represents the rate of change of volume, which is given as 11 in³/min. So, we have:

`\Longrightarrow (dx)/(dt)=(\Big((dV)/(dt)\Big) )/(\Big(\frac43 \pi\big[3r^2+6rx+3x^2]\Big))\\\\\\\\\therefore \boxed{\boxed{(dx)/(dt)=(3\Big((dV)/(dt)\Big) )/(4 \pi\big(3r^2+6rx+3x^2\big))}}`

Part (c): To find the rate at which the thickness of the ice is decreasing when it is 3 inches thick, we substitute x = 3, r=6, and dV/dt=-11 ("-" since it is melting) into the equation and solve for dx/dt.

`(dx)/(dt)=(3\Big((dV)/(dt)\Big) )/(4 \pi\big(3r^2+6rx+3x^2\big))\\\\\\\\\Longrightarrow (dx)/(dt)=(3(-11) )/(4 \pi\big(3(6)^2+6(6)(3)+3(3)^2\big))\\\\\\\\\Longrightarrow (dx)/(dt)=-(33 )/(4 \pi(243))\\\\\\\\\Longrightarrow (dx)/(dt)=-(33 )/(972 \pi)\\\\\\\\\therefore \boxed{\boxed{(dx)/(dt)=-(11 )/(324 \pi) \ \text{in}/\text{min}}} \ \text{or} \ \boxed{\boxed{(dx)/(dt) \approx -0.0108 \ \text{in}/\text{min}}}`

Thus, the rate at which the thickness of the ice is decreasing is 0.0108 in/min.

Part (d): The surface area of the sphere, "A," can be expressed as:

`\rightarrow A=4\pi r^2`

Where "r" is the radius of the sphere.

The outer surface area of the ice can be expressed as:

`\rightarrow \boxed{\boxed{A=4\pi(r + x)^2}}`

Where "r" is the radius of the sphere and "x" is the thickness of the ice.

Part (e): Let's differentiate both sides of the equation with respect to t:

`A=4\pi(r + x)^2\\\\\\\\\therefore \boxed{\boxed{ (dA)/(dt)=8 \pi (r+x) \cdot (dx)/(dt) }}`

Part (f): To find the rate at which the outer surface area of the ice is decreasing when it is 3 inches thick, we substitute x = 3, r=6, and dx/dt=-0.0108 into the equation and solve for dA/dt.

`(dA)/(dt)=8 \pi (r+x) \cdot (dx)/(dt) \\\\\\\\\Longrightarrow (dA)/(dt)=8 \pi (6+3) \cdot (-0.0108)\\\\\\\\\therefore \boxed{\boxed{(dA)/(dt) \approx -2.44 \ \text{in}^2/ \text{min} }}`

Thus, the rate at which the outer surface area is decreasing is 2.44 in²/min.