Problem 41 - Calculus for Biology and Medicine

Question

Differentiate $$ R(T)=\frac{2 \pi^{5}}{15} \frac{k^{4}}{c^{2} h^{3}} T^{4} $$

Step-by-Step Solution

Answer

The derivative of \( R(T) \) with respect to \( T \) is \( \frac{8 \pi^{5}}{15} \frac{k^{4}}{c^{2} h^{3}} T^{3} \).

Step 1: Identify the Function to Differentiate

The function we are tasked to differentiate is given as \[ R(T) = \frac{2 \pi^{5}}{15} \frac{k^{4}}{c^{2} h^{3}} T^{4} \]where \( T \) is the variable, and the other quantities are constants.

Step 2: Apply the Power Rule for Differentiation

The function can be simplified for differentiation as \[ R(T) = C \cdot T^{4} \]where \( C = \frac{2 \pi^{5}}{15} \frac{k^{4}}{c^{2} h^{3}} \) is a constant. To differentiate \( R(T) \) with respect to \( T \), the power rule is applied: \( \frac{d}{dT} [T^n] = n T^{n-1} \).

Step 3: Differentiate Step-by-Step

Now apply the power rule:\[ \frac{dR}{dT} = \frac{d}{dT}[\, C \cdot T^{4}\,] = C \cdot 4 \cdot T^{4-1} = 4C \cdot T^{3} \]where we have multiplied the constant \( C \) by the derivative of \( T^4 \).

Step 4: Substitute Back the Constant

Substitute the expression for \( C \) back into the derivative:\[ \frac{dR}{dT} = 4 \left( \frac{2 \pi^{5}}{15} \frac{k^{4}}{c^{2} h^{3}} \right) T^{3} = \frac{8 \pi^{5}}{15} \frac{k^{4}}{c^{2} h^{3}} T^{3} \]

Key Concepts

Power Rule

The power rule is a foundational concept in differentiation that simplifies the process of finding derivatives for functions with variables raised to a power. This rule states that if you have a function in the form of \( T^n \), where \( T \) is your variable and \( n \) is a real number, the derivative is obtained by multiplying the exponent by the variable raised to one less than the original exponent. In algebraic form: \( \frac{d}{dT} [T^n] = n T^{n-1} \).

Understanding the power rule is essential as it allows for quick computation of derivatives, especially for polynomial terms. In our exercise, we encountered \( T^{4} \). By using the power rule, we determined the derivative as \( 4 T^{3} \).

This method is a time-saver compared to calculating limits, and it applies not only to simple monomials but to more complex expressions where parts of the function include powers of variables.

Constants in Calculus

In calculus, constants play a crucial role, especially when it comes to differentiation. A constant is simply a fixed value that does not change as the variables do. When a function is differentiated, the constants act as multipliers and have no impact on the operation of differentiation itself.

For our function \( R(T) = C \cdot T^4 \), we designated \( C = \frac{2 \pi^{5}}{15} \frac{k^{4}}{c^{2} h^{3}} \) as the constant. During the differentiation process, this constant can be factored out of the derivative. Therefore, it simplifies our calculations since we only need to apply differentiation to the variable term, in this case, \( T^4 \). This practice shows the beauty of calculus in making complex expressions simpler and more manageable.

• Constants do not change with the variable
• They can be factored out during differentiation
• Simplify expressions by reducing unnecessary computations

Variable Differentiation

Differentiation with respect to a variable means finding how one quantity changes as another quantity changes. This process is important in many fields such as physics and engineering, where understanding the rate of change is necessary to describe natural phenomena.

In the context of the original exercise, differentiating \( R(T) \) with respect to \( T \) requires looking at how \( R(T) \) changes as \( T \) changes. Given that \( T \) is the only variable and all other terms are constants, the focus is solely on determining the derivative \( \frac{dR}{dT} \).

Differentiating solely with respect to \( T \) highlights how crucial isolating the variable of interest is. This allows for simplification of the problem without unnecessary complexity. In practice, always isolate the variable with respect to which you are performing differentiation, as it allows for clearer insights and more efficient computations.