Problem 11 - Essential Calculus

Question

Find any critical numbers of the function. $$ g(t)=t \sqrt{4-t}, t<3 $$

Step-by-Step Solution

Answer

The critical number for the function \(g(t) = t \cdot \sqrt{4 - t}\) for \(t < 3\) is \(t = \frac{8}{3}\).

Step 1: Applying the Product Rule for Derivatives.

First differentiate the given function. Since g(t) is a product of two functions, we will apply the product rule of derivates which states: if a function \(h(t) = f(t) \cdot g(t)\), then its derivative is given by \(h'(t) = f'(t) \cdot g(t) + f(t) \cdot g'(t)\). Applying this rule to the given function \(g(t) = t \cdot \sqrt{4-t}\), we obtain:\(g'(t) = 1 \cdot \sqrt{4-t} + t \cdot \frac{1}{2\sqrt{4-t}}(-1)\)

Step 2: Simplifying the first derivative and finding critical points.

Simplify the derivative as: \(g'(t) = \sqrt{4-t} - \frac{t}{2 \sqrt{4-t}}\). The critical numbers are those \(t\) values that make \(g'(t) = 0\) or the values that make \(g'(t)\) undefined. Solve the equation \(g'(t) = 0\), by first multiplying both sides by \(2 \sqrt{4-t}\) to get: \(0 = 2(4-t) - t = 8 - 3t\). Solving this for \(t\), we get: \(t = \frac{8}{3}\). The derivative is undefined at \(t=4\) but because this value is outside the given interval (t<3), we disregard it. So, \(t = \frac{8}{3}\) is the only critical number exists on our interval.

Step 3: Verifying the validity of the critical number.

Verify whether the derived critical number falls within the allocated interval. Since \(t = \frac{8}{3}\) is less than 3, it is a valid critical number for the given function. Thus, we conclude that the function \(g(t) = t \cdot \sqrt{4 - t}\) for \(t < 3\) has one critical number at \(t = \frac{8}{3}\).

Key Concepts

Product Rule

When dealing with derivatives of a product of two functions, the Product Rule is a valuable tool. Imagine you have a function that consists of two multiplied parts, say \(h(t) = f(t) \cdot g(t)\). To find the derivative \(h'(t)\), you cannot simply differentiate each part separately. Instead, you need the Product Rule, which states:

• \(h'(t) = f'(t) \cdot g(t) + f(t) \cdot g'(t)\).

In the given exercise, \(g(t) = t \cdot \sqrt{4-t}\), we treat \(t\) as \(f(t)\) and \(\sqrt{4-t}\) as \(g(t)\). Deriving each part separately and applying the rule, we get:

• The derivative of \(t\) is simply 1.
• The derivative of \(\sqrt{4-t}\) using the chain rule is \(-\frac{1}{2\sqrt{4-t}}\).

So, by applying the Product Rule, the derivative turns into \(\sqrt{4-t} - \frac{t}{2\sqrt{4-t}}\). This allows us to explore changes in our function.

Derivative

Derivatives help us understand how a function changes at any given point. They provide the rate of change and can tell us important features about the function like increasing or decreasing behavior.
In this exercise, to find the critical numbers, we first need to determine the derivative \(g'(t)\) of \(g(t) = t \cdot \sqrt{4-t}\). With the calculated derivative \(g'(t) = \sqrt{4-t} - \frac{t}{2\sqrt{4-t}}\), we analyze how \(g(t)\) behaves:

• If \(g'(t) > 0\), the function is increasing at \(t\).
• If \(g'(t) < 0\), the function is decreasing at \(t\).
• If \(g'(t) = 0\), \(t\) might be a critical point, where changes in direction can occur (such as peaks or troughs).

By simplifying the derivative to \(g'(t) = \sqrt{4-t} - \frac{t}{2\sqrt{4-t}}\), we can locate critical points by solving when this equals zero, highlighting where the function could potentially have peaks or changes.

Critical Points

Critical points of a function are where the derivative is zero or undefined, indicating potential maxima, minima, or saddle points. These are essential for understanding the overall behavior and key characteristics of the function.
For our function \(g(t) = t \cdot \sqrt{4-t}\), we found \(g'(t) = \sqrt{4-t} - \frac{t}{2\sqrt{4-t}}\). We need to solve for \(t\) when \(g'(t) = 0\). After simplifying:

• Multiply both sides of the equation by \(2\sqrt{4-t}\) to clear the fraction.
• You end up with \(0 = 8 - 3t\).
• Solving gives \(t = \frac{8}{3}\).

This value, \(t = \frac{8}{3}\), is within the original constraint \(t < 3\), thus making it a valid critical number. It indicates a point where the rate of change of the function shifts, possibly marking a local extremum.