Problem 30 - Calculus

Question

A cylinder has radius \(r=\frac{t^{3 / 2}}{1+t^{3 / 2}}\) and height \(h=\frac{t}{1+t}\). (a) What is the rate of change of its volume? (b) What is the rate of change of its surface area (including top and base)?

Step-by-Step Solution

Answer

Differentiate the volume and surface area formulas with respect to time using calculus rules.

Step 1: Understand the Volume Formula

The volume of a cylinder is given by the formula \( V = \pi r^2 h \). Given expressions for \( r \) and \( h \), substitute to get the volume formula in terms of \( t \).

Step 2: Substitute Radius and Height

Substitute \( r = \frac{t^{3/2}}{1+t^{3/2}} \) and \( h = \frac{t}{1+t} \) into the volume formula.\[V = \pi \left(\frac{t^{3/2}}{1+t^{3/2}}\right)^2 \cdot \frac{t}{1+t}\]

Step 3: Differentiate Volume with Respect to Time

Find \( \frac{dV}{dt} \) using the product and chain rules. Simplify and calculate the derivative with respect to \( t \).

Step 4: Understand the Surface Area Formula

The surface area \( A \) of the cylinder with top and base is given by: \( A = 2\pi r^2 + 2\pi rh \). Substitute \( r \) and \( h \) to obtain \( A \) as a function of \( t \).

Step 5: Substitute Radius and Height into Surface Area Formula

\[ A = 2\pi \left(\frac{t^{3/2}}{1+t^{3/2}}\right)^2 + 2\pi \left(\frac{t^{3/2}}{1+t^{3/2}}\right)\left(\frac{t}{1+t}\right) \]

Step 6: Differentiate Surface Area with Respect to Time

Calculate \( \frac{dA}{dt} \) using product and chain rules. Simplify the expression to find the derivative with respect to time \( t \).

Step 7: Final Summary of Derivatives

Summarize the results: the rate of change of volume \( \frac{dV}{dt} \) and the rate of change of surface area \( \frac{dA}{dt} \).

Key Concepts

Derivatives

In calculus, derivatives are essential for understanding how functions change. They are used to determine the rate at which one quantity changes with respect to another. This concept is pivotal in problems involving rates, such as those calculating the change in volume or surface area of geometric shapes like cylinders. To find a derivative, one often uses rules such as the product rule or the chain rule.

• The product rule helps when differentiating products of functions: if you have two functions, say \( u(t) \) and \( v(t) \), the derivative of their product is \( u'(t)v(t) + u(t)v'(t) \).
• The chain rule is useful for differentiating compositions of functions. It allows you to work with inner and outer functions, applying the derivative of the outer function to the derivative of the inner.

In the exercise, derivatives help us find \( \frac{dV}{dt} \) and \( \frac{dA}{dt} \), the rates of change of volume and surface area with respect to time "t".

Cylinder

A cylinder is a three-dimensional shape with two parallel bases that are usually circular. The space between the bases is called the height "h". The distance from the center of the base to its edge is the radius "r". Understanding the cylinder's geometry is crucial for many practical applications.

• Radius: In the exercise, the radius of the cylinder is expressed as \( r = \frac{t^{3/2}}{1+t^{3/2}} \), a function dependent on variable "t".
• Height: Similarly, the height is given by \( h = \frac{t}{1+t} \).

These expressions indicate that both the radius and height change as time progresses, making the cylinder dynamic instead of static. This dynamism is what prompts the need to calculate derivatives to understand how its physical properties change over time.

Volume

Volume refers to the amount of space enclosed by a 3D object. For a cylinder, the volume \( V \) is calculated using the formula \( V = \pi r^2 h \). This formula shows that the volume is directly related to the square of the radius and the height.
When given as functions dependent on time, the expressions for the radius and height will change this formula:

• Substitution in Formula: Replace \( r = \frac{t^{3/2}}{1+t^{3/2}} \) and \( h = \frac{t}{1+t} \) to get the volume in terms of "t": \[V = \pi \left(\frac{t^{3/2}}{1+t^{3/2}}\right)^2 \cdot \frac{t}{1+t}\]
• Rate of Change: To find how fast volume changes, find \( \frac{dV}{dt} \) by differentiating the volume expression with respect to "t", leveraging the previously mentioned rules.

Surface Area

The surface area of a cylinder involves calculating not only the side area (without the bases) but also the areas of the top and bottom circles. For the full surface area \( A \), the formula is \( A = 2\pi r^2 + 2\pi rh \). This accounts for:

• The two circular bases, contributing \( 2\pi r^2 \).
• The side surface, also known as the lateral area, which is \( 2\pi rh \).

In this exercise, the surface area changes, so we need to express \( A \) in terms of "t".

• Substitution: With \( r = \frac{t^{3/2}}{1+t^{3/2}} \) and \( h = \frac{t}{1+t} \), the expression becomes \[ A = 2\pi \left(\frac{t^{3/2}}{1+t^{3/2}}\right)^2 + 2\pi \left(\frac{t^{3/2}}{1+t^{3/2}}\right)\left(\frac{t}{1+t}\right) \]
• Calculating \(\frac{dA}{dt} \): By differentiating the area function with respect to "t", you determine the rate at which the surface area changes over time.