Question

Volume and Surface Area The radius of a right circular cylinder is increasing at a rate of 6 inches per minute, and the height is decreasing at a rate of 4 inches per minute. What are the rates of change of the volume and surface area when the radius is 12 inches and the height is 36 inches?

Short answer

The volume of the cylinder is changing at a rate of \( 4608 \pi \) in³/min and the surface area is changing at a rate of \( 480 \pi \) in²/min.

Step-by-step solution

Identify the Equations for Volume and Surface Area of a Cylinder

The volume \(V\) of a right circular cylinder with radius \(r\) and height \(h\) is \(V = \pi r^2 h\) and the total surface area \(A\) is \(A = 2 \pi r h + 2 \pi r^2\).

Differentiate the Volume and Surface Area with respect to time

Differentiate both sides of the volume equation with respect to \(t\), \(\frac{dV}{dt} = 2\pi r \frac{dr}{dt}h + \pi r^2 \frac{dh}{dt}\), and the surface area equation, \(\frac{dA}{dt} = 2 \pi \frac{dr}{dt}h + 2 \pi r \frac{dh}{dt} + 2 \pi r \frac{dr}{dt}\).

Substitute the given values

Substitute the given values where the rate of change of radius \( \frac{dr}{dt} = 6 \) in/min, rate of change of height \( \frac{dh}{dt} = -4 \) in/min, the radius \( r = 12 \) in and the height \( h = 36 \) in into the differentiated equations. So for volume equation we get, \(\frac{dV}{dt} = 2\pi * 12 * 6 * 36 + \pi * (12)^2 * (-4) = 5184 \pi - 576 \pi = 4608 \pi \) in³/min and for surface area equation, \(\frac{dA}{dt} = 2\pi * 6 * 36 + 2 \pi * 12 * -4 + 2 \pi * 12 * 6 = 432 \pi - 96 \pi + 144 \pi = 480 \pi\) in²/min. Thus, the volume and surface area of the cylinder are changing at rates of \( 4608 \pi \) in³/min and \( 480 \pi \) in²/min, respectively.

Key concepts

Cylinder Volume and Surface Area

Understanding the volume and surface area of a cylinder is fundamental in many areas of mathematics and physics. The volume of a cylinder can be found using the formula:
\( V = \text{π}r^2h \),
where \(r\) represents the radius and \(h\) the height of the cylinder. Surface area, on the other hand, combines the areas of the two circular bases and the rectangular wrap around the side. The surface area is given by:
\( A = 2\text{π}rh + 2\text{π}r^2 \).

When determining how these values change over time, as with the cylinder whose dimensions are not static, we monitor the rates of change of the radius and height, and through this, infer the rates at which the volume and surface area evolve. This contextual understanding helps bring the abstract formulas to life and shows their practical applications, such as in mechanical engineering where parts may expand or contract under heat.

Differentiation of Equations

Differentiation is a powerful tool in calculus to find how one quantity changes with respect to another. For related rates problems involving a cylinder, we differentiate the volume and surface area equations with respect to time. The derivative of these equations reveals the relationship between the change in the cylinder’s dimensions and the rate at which the volume and surface area change.

In our example, we differentiated the volume and surface area equations as:
\( \frac{dV}{dt} = 2\text{π}r\frac{dr}{dt}h + \text{π}r^2\frac{dh}{dt} \) and \( \frac{dA}{dt} = 2\text{π}h\frac{dr}{dt} + 2\text{π}r\frac{dh}{dt} + 2\text{π}r\frac{dr}{dt}\).
This process, known as implicit differentiation, allows us to calculate the instantaneous rates of change (in this scenario, the rates of change of the volume and surface area of the cylinder), an invaluable analysis for interpreting dynamic physical systems.

Applying Rates of Change

Applying rates of change in calculus often involves relating varying quantities through their rates. After getting the derivatives, as shown in previous sections, we need to substitute known rates and values to find the unknown rates. For the cylinder's changing volume and surface area:

We substitute the rates \(\frac{dr}{dt}=6\) inches/minute and \(\frac{dh}{dt}=-4\) inches/minute, and the instantaneous values of radius \(r=12\) inches and height \(h=36\) inches. Following substitution, we solve for \(\frac{dV}{dt}\) and \(\frac{dA}{dt}\), which are the rates of change for volume and surface area, respectively.

In practical situations, such as monitoring the water level in a tank or the expansion of materials under various conditions, understanding and applying rates of change allow us to predict and manage physical changes effectively. Calculus provides us with the essential framework for making these calculations, linking seemingly separate concepts through a common mathematical language.