Problem 7 - Applied Calculus

Question

Find the partial derivatives in Problems. The variables are restricted to a domain on which the function is defined. $$f_{t} \text { if } f(t, a)=5 a^{2} t^{3}$$

Step-by-Step Solution

Answer

The partial derivative \( f_t = 15a^2 t^2 \).

Step 1: Understand the Function

We are given a function \( f(t, a) = 5a^2 t^3 \). The task is to find the partial derivative of this function with respect to \( t \).

Step 2: Identify the Variable for Differentiation

In this problem, we are finding the partial derivative with respect to \( t \). Thus, treat \( a \) as a constant.

Step 3: Apply the Power Rule

The power rule for differentiation states that if \( x^n \), the derivative with respect to \( x \) is \( nx^{n-1} \). Apply this rule to the term \( t^3 \).

Step 4: Differentiate the Function

Differentiate \( f(t, a) = 5a^2 t^3 \) with respect to \( t \). Treat \( 5a^2 \) as a constant. The derivative is \( 3 \times 5a^2 t^{3-1} = 15a^2 t^2 \).

Step 5: Write the Result

The partial derivative of the function with respect to \( t \) is \( f_t = 15a^2 t^2 \).

Key Concepts

Power Rule

One fundamental tool in calculus, particularly in differentiation, is the power rule. This rule simplifies the process of finding derivatives for power functions of the form \(x^n\). It states that to differentiate \(x^n\), you simply multiply by the exponent \(n\) and reduce the power by one. So, the derivative of \(x^n\) is \(nx^{n-1}\).

In the context of differentiating a multivariable function such as \(f(t, a) = 5a^2 t^3\), the power rule only applies to the variable you are differentiating with respect to, in this case, \(t\).
When applying this rule to \(t^3\), you derive \(3t^{3-1} = 3t^2\). Therefore, the derivative with respect to \(t\) is \(3 \cdot 5a^2 \cdot t^2\), which simplifies to \(15a^2 t^2\).

Using the power rule is efficient and essential in calculus. It allows us to easily find the derivative of polynomial functions, which are common in mathematical models and real-world applications.

Function Differentiation

Differentiation is a core concept in calculus and is used to determine the rate at which a function is changing at any given point. It involves computing the derivative of a function, which is essentially finding an expression for the rate of change.

The function \(f(t, a) = 5a^2 t^3\) is a product of terms that include the variable \(t\) raised to a power and a constant term \(5a^2\). When differentiating with respect to \(t\), you keep the constants \(5\) and \(a^2\) unchanged and differentiate only the \(t^3\) term.

• Identify which variable you are differentiating in terms of. In this exercise, we differentiate with respect to \(t\).
• Keep all other terms that don't involve \(t\) as constants while applying differentiation rules like the power rule.

By finding the derivative, you know how the function \(f(t, a)\) changes as \(t\) changes, which is particularly useful in understanding dynamic systems and optimization problems.

Multivariable Functions

Multivariable functions depend on more than one variable, like \(f(t, a)\) which depends on both \(t\) and \(a\). When dealing with such functions, partial derivatives are used to differentiate one variable, treating others as constants.

To find the partial derivative of a multivariable function, you choose one variable to differentiate with respect to (partial differentiation), and fix all other variables as constants.

• In \(f(t, a) = 5a^2 t^3\), find the partial derivative \(f_t\), treating \(a\) as a constant.
• This means applying differentiation rules only to \(t\) while leaving the \(a\)-related terms untouched.

Partial derivatives allow us to explore how a function behaves as one specific variable changes. This is critical in multivariable calculus, where functions model complex relationships involving multiple changing quantities, and in fields such as physics, engineering, and economics.