Problem 17 - Calculus Early Transcendentals

Question

The volume of a right circular cylinder with radius \(r\) and height \(h\) is \(V=\pi r^{2} h\). a. Assume that \(r\) and \(h\) are functions of \(t .\) Find \(V^{\prime}(t)\). b. Suppose that \(r=e^{t}\) and \(h=e^{-2 t},\) for \(t \geq 0 .\) Use part (a) to find \(V^{\prime}(t)\). c. Does the volume of the cylinder in part (b) increase or decrease as \(t\) increases?

Step-by-Step Solution

Answer

Answer: The volume of the cylinder is neither increasing nor decreasing as t increases, as the derivative V'(t) = 0.

Step 1: Part a: Find \(V'(t)\) in terms of \(r'(t)\) and \(h'(t)\)

Start by applying the Chain Rule to obtain the derivative of the volume equation, \(V'(t)\). We have: $$ \frac{dV}{dt} = \frac{d(\pi r^2h)}{dt} $$ Now, by using the product rule, we can rewrite the above expression as: $$ \frac{dV}{dt} = \pi \left(\frac{d(r^2h)}{dt}\right) = \pi\left(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\right) $$ So, the expression for \(V'(t)\) in terms of \(r'(t)\) and \(h'(t)\) is: $$ V'(t) = \pi(2r(t)r'(t)h(t) + r^2(t)h'(t)) $$

Step 2: Part b: Find \(V'(t)\) for \(r=e^{t}\) and \(h=e^{-2t}\)

Given that \(r(t)=e^t\) and \(h(t)=e^{-2t}\), we first need to find \(r'(t)\) and \(h'(t)\). By taking the derivatives with respect to \(t\) for each function, we get: $$ r'(t) = \frac{d(e^t)}{dt} = e^t $$ and $$ h'(t) = \frac{d(e^{-2t})}{dt} = -2e^{-2t} $$ Now, we can substitute these values, along with \(r(t)\) and \(h(t)\), into the expression for \(V'(t)\) from part a: $$ V'(t) = \pi(2e^{t}e^t \cdot e^{-2t} + e^{2t} \cdot (-2e^{-2t})) = \pi(2e^0 - 2e^0) = \pi(2-2) = 0 $$ So, \(V'(t) = 0\) for the given expressions of \(r(t)\) and \(h(t)\).

Step 3: Part c: Determine if the volume of the cylinder is increasing or decreasing

Since the derivative \(V'(t) = 0\), there is no change in the volume of the cylinder as \(t\) increases. The volume is neither increasing nor decreasing.

Key Concepts

Chain Rule

When dealing with derivatives in calculus, the Chain Rule is essential. It allows us to find the derivative of a composition of functions. Imagine you have a function that is inside another function, like a nested set of Russian dolls. For example, if you know that the quantity you're interested in, such as the volume of a cylinder, depends on time through several steps, the Chain Rule helps you differentiate this complex relationship.

In our exercise, the volume of the cylinder, \(V\), is given by \(\pi r^2 h\). Both \(r\) and \(h\) are functions of time \(t\). To find how the volume changes with time, \(V'(t)\), we apply the Chain Rule since both the radius \(r\) and height \(h\) change with time but are not explicitly given by a simple function of \(t\). This leads us to an equation that delineates the relationship between the derivative of the volume with respect to time and the derivatives of radius and height. By applying the Chain Rule properly, we accurately grasp the dynamic nature of the volume as affected by time.

Product Rule

The Product Rule is another fundamental concept in calculus. It deals with differentiating products of two or more functions. In simple terms, when you have a function that is the product of two separate functions, the derivative cannot just be the product of the derivatives. Instead, you use the Product Rule. It is expressed in a formulaic way: if you have two functions, \(u\) and \(v\), then their derivative \( (uv)' \) is \( u'v + uv' \).

• \(u'v\) is the derivative of the first function times the second function,
• and \(uv'\) is the first function times the derivative of the second function.

When applied to our cylinder volume problem, the Product Rule helps in finding the derivative of \( V = \pi r^2 h \). By breaking it into parts, we apply the Product Rule to \(r^2\) and \(h\), separately. It lets us see how changes in either the radius or the height, independently, affect the rate of volume change. The holistic nature of the Product Rule allows capturing intricate relationships and dependencies that simpler differentiation rules might miss.

Exponential Functions

Exponential functions are a unique class of functions, often characterized by constant ratios of growth. These functions usually take the form \(a^x\), where \(a\) is a constant and \(x\) is the exponent. They have a distinct property: their rate of change is proportional to their current value, making them extremely useful for modeling growth and decay situations—from population growth to radioactive decay.

In the context of our exercise, the radius \(r\) and height \(h\) of the cylinder are each given as exponential functions of time \(t\): \(r(t) = e^t\) and \(h(t) = e^{-2t}\). When finding their derivatives, you see both behave predictably: the derivative of \(e^t\) is \(e^t\) (a unique feature of the natural exponential function), and the derivative of \(e^{-2t}\) is \(-2e^{-2t}\).
These insights into the nature of exponential functions allow us to perceive how variables impacted by exponential growth or decay can influence other dependent variables like the volume of the cylinder. Understanding this dynamic interaction can be particularly powerful when dealing with real-world phenomena.