Question

Find the derivative with respect to the independent variable. $$s=58.3 t^{3}-63.8 t$$

Short answer

The derivative of \( s = 58.3t^3 - 63.8t \) with respect to t is \( s' = 174.9t^2 - 63.8 \).

Step-by-step solution

Identify the function to differentiate

The function given is in terms of the independent variable t, which is \( s = 58.3t^3 - 63.8t \). We need to find the derivative of this function with respect to t.

Apply the power rule to each term

To differentiate the function, use the power rule. The power rule states that the derivative of \( t^n \) with respect to t is \( n \cdot t^{n-1} \). Apply this to each term independently: - The derivative of \( 58.3t^3 \) with respect to t is \( 3 \cdot 58.3t^{3-1} = 174.9t^2 \).- The derivative of \( -63.8t \) with respect to t (treating \( t \) as \( t^1 \)) is \( 1 \cdot -63.8 \cdot t^{1-1} = -63.8 \).

Combine the derivatives

By combining the derivatives of the two terms, we get the complete derivative of the function s with respect to t, which is \( s' = 174.9t^2 - 63.8 \).

Key concepts

Power Rule Differentiation

Understanding the power rule is essential when it comes to differentiating polynomials. It's a straightforward technique that significantly simplifies the process of taking derivatives. The power rule states that for any function of the form
\( f(t) = at^n \),
where
\( a \) and
\( n \) are constants, and
\( t \) is the variable, the derivative of
\( f \) with respect to
\( t \) is
\( f'(t) = a \times n \times t^{n-1} \).

To apply this to a polynomial, differentiate each term independently before combining them to get the overall derivative. This makes the power rule a key tool for finding the rate of change, growth, and solving many real-world problems. It’s particularly useful because it doesn’t require us to expand binomials or perform any complex algebraic manipulations — a time saver for sure!

Derivative of a Function

The derivative of a function is a fundamental concept in calculus that measures how a function changes as its input changes. It represents the instantaneous rate of change or the slope of the tangent line at any point on the function's graph. The process of calculating a derivative is called differentiation. For our function
\( s=58.3t^3-63.8t \),
the derivative
\( s' \) tells us how the value of
\( s \) changes with respect to changes in
\( t \).

In practical terms, if
\( t \) represented time, and
\( s \) represented distance, then
\( s' \) would give the speed at any given moment. Understanding the derivative is vital not only in mathematics but in various fields including physics, engineering, and economics, as it provides detailed insights into the dynamics of systems and helps to predict future behavior.

Independent Variable in Calculus

In calculus, the independent variable is the value that we can manipulate directly, and it's often represented as
\( x \) or
\( t \).
It's the 'input' for functions, which becomes especially important when discussing derivatives. When we differentiate a function with respect to an independent variable, we are interested in finding out how the function's output (dependent variable) changes as the independent variable changes.

In our exercise, the independent variable is
\( t \), which might represent time in a given context — although it could stand for any quantity that can vary independently. The proper understanding of independent variables allows us to correctly set up functions for modeling real-life scenarios and predict how changing one quantity affects another, which is at the heart of differential calculus.