Problem 15 - Applied Calculus

Question

If \(f(u, v)=5 u v^{2}\), find \(f(3,1), f_{u}(3,1)\), and \(f_{v}(3,1)\).

Step-by-Step Solution

Answer

\(f(3, 1) = 15\), \(f_u(3, 1) = 5\), \(f_v(3, 1) = 30\).

Step 1: Substituting into the Function

To find \(f(3,1)\), substitute \(u = 3\) and \(v = 1\) into the function \(f(u, v) = 5uv^2\). This gives \(f(3, 1) = 5 \times 3 \times 1^2\). Calculate the expression.

Step 2: Calculating the Result

Perform the calculation from the previous step: \(f(3, 1) = 5 \times 3 \times 1 = 15\). So, the value of the function at the point \((3, 1)\) is 15.

Step 3: Partial Derivative with Respect to u

Differentiate the function \(f(u, v) = 5uv^2\) with respect to \(u\). The partial derivative is \(f_u(u, v) = 5v^2\), as the derivative of \(u\) with respect to itself is 1, and \(v^2\) is treated as a constant multiplier.

Step 4: Evaluating the Partial Derivative f_u(3,1)

Substitute \(v = 1\) into \(f_u(u, v) = 5v^2\). This gives \(f_u(3, 1) = 5 \times 1^2 = 5\). So, \(f_u(3, 1) = 5\).

Step 5: Partial Derivative with Respect to v

Differentiate the function \(f(u, v) = 5uv^2\) with respect to \(v\). Here, use the power rule for \(v^2\), giving \(f_v(u, v) = 10uv\) because \(2v = \frac{d}{dv}\ (v^2)\).

Step 6: Evaluating the Partial Derivative f_v(3,1)

Substitute \(u = 3\) and \(v = 1\) into \(f_v(u, v) = 10uv\). This yields \(f_v(3, 1) = 10 \times 3 \times 1 = 30\). Therefore, \(f_v(3, 1) = 30\).

Key Concepts

Partial Derivatives

In multivariable calculus, partial derivatives are used to understand how the function changes as only one of its variables changes at a time, while keeping all other variables constant. This is especially useful for functions with more than one variable because we can identify the effect of changing each variable independently.
To compute a partial derivative for a function like \( f(u, v) = 5uv^2 \), we choose one variable to differentiate with respect to, treating other variables as constants.

• Partial Derivative with Respect to \( u \): When differentiating with respect to \( u \), the term \( v^2 \) is held constant. Thus, the partial derivative \( f_u(u, v) = 5v^2 \).
• Partial Derivative with Respect to \( v \): While differentiating with respect to \( v \), we apply the power rule to \( v^2 \) treating \( u \) as a constant. Therefore, the partial derivative becomes \( f_v(u, v) = 10uv \).

Function Evaluation

Function evaluation involves plugging specific values into a given multivariable function to find the output. It helps to identify the behavior of the function at particular points.
For the function \( f(u, v) = 5uv^2 \), if we want to find \( f(3, 1) \), we substitute \( u = 3 \) and \( v = 1 \). This means we calculate \( f(3,1) = 5 \times 3 \times 1^2 \). By simplifying, we get \[ f(3,1) = 15 \]. Consequently, the value of the function at the point \((3,1)\) is \( 15 \).

• Substitution involves replacing each variable in the function's expression with given numerical values.
• This step is crucial for understanding specific outputs and aiding in verifying solution correctness.

Power Rule

The power rule is a basic yet essential concept in calculus utilized frequently when finding derivatives. It's a quick way to get the derivative of an expression of the form \( x^n \), where \( n \) is any real number. According to the power rule, the derivative of \( x^n \) is \( nx^{n-1} \).
For the function \( f(u, v) = 5uv^2 \), the power rule is used when differentiating with respect to \( v \).

• When we differentiate \( v^2 \), we get \( 2v \) because of the power rule: \( \frac{d}{dv}(v^2) = 2v \).
• Applying the power rule helps simplify the differentiation process and provides accurate results swiftly.

In this context, applying the power rule gives us the partial derivative \( f_v(u, v) = 10uv \), since \( 10u \) multiplies the derivative of \( v^2 \). This step highlights how foundational the power rule is in handling derivatives in multivariable calculus.