Question

Find the derivative with respect to the independent variable. $$v=5 t^{2}-3 t+4$$

Short answer

\(v'(t) = 10t - 3\)

Step-by-step solution

Identify the Function

Identify the given function. In this case, the given function is the velocity, expressed as a polynomial of time: \(v(t) = 5t^2 - 3t + 4\).

Apply the Power Rule

For each term in the function, apply the power rule of differentiation, \(\frac{d}{dt} [t^n] = n \cdot t^{n-1}\). Apply this rule to each term separately.

Differentiate the First Term

Differentiate the first term \(5t^2\), which using the power rule becomes \(2 \cdot 5t^{2-1} = 10t\).

Differentiate the Second Term

Differentiate the second term \(-3t\), which is a linear term and its derivative is the coefficient of \(t\), thus \(-3\).

Differentiate the Third Term

Differentiate the constant term \(+4\), which has a derivative of zero since the derivative of any constant is zero.

Combine the Derivatives

Combine the derivatives of all terms to obtain the derivative of the entire function with respect to time.

Key concepts

Power Rule Differentiation

Understanding the power rule for differentiation is crucial for solving calculus problems involving derivatives. This rule states that if you have a function in the form of a monomial, which is a single term of the form \(c \cdot t^n\), where \(c\) is a constant and \(n\) is a real number exponent, the derivative of that monomial with respect to the variable \(t\) is \(n \cdot c \cdot t^{n-1}\). Essentially, you bring down the exponent as a coefficient and subtract one from the original exponent.

For example, if you want to differentiate \(5t^2\), you would multiply the exponent 2 by the coefficient 5 to get 10, and then decrease the exponent by one, resulting in \(10t^1\) or simply \(10t\). This step simplifies the process of differentiation significantly, making it easier to tackle even more complex functions involving polynomials and to quickly determine the rate of change at any given point.

Polynomial Differentiation

Polynomials are algebraic expressions consisting of multiple terms, each a constant multiplied by a variable raised to a non-negative integer exponent, like \(5t^{2}-3t+4\). When differentiating a polynomial, we apply the power rule to each term separately, as polynomials are sums of monomials.

The differentiation of a polynomial results in another polynomial where each term's degree (the exponent) is reduced by one. A critical point to note is that the derivative of a constant term (a term with no variable, such as \(+4\)) is always zero because constants do not change and, hence, have no rate of change. After computing the derivatives of individual terms, we combine them to get the derivative of the entire polynomial, which in the context of calculus reflects a number of real-life phenomena such as velocity and acceleration in terms of time.

Velocity Function Calculus

In calculus, velocity is a function that describes an object's rate of change of position with respect to time. It is a vector quantity, meaning it has both magnitude and direction, but in many one-dimensional problems, it's treated simply as a rate (speed with direction) over time. The velocity function can often be expressed as a polynomial of time, resembling the form \(v(t) = at^{n} + ... + bt + c\), where each term represents different components of motion.

When finding the derivative of the velocity function, like \(v(t) = 5t^{2} - 3t + 4\), we're effectively calculating the acceleration function; it tells us how the velocity is changing over time. As demonstrated in the exercise, we differentiate each term of the velocity polynomial to find this acceleration function. Hence, understanding how to differentiate polynomials, especially through the power rule, is paramount for analyzing motion and predicting future behavior of moving objects under various forces.