Problem 23 - Calculus an Applied Approach

Question

find the given value. $$ g(t)=5 t^{4}+10 t^{2}+3 \quad g^{\prime \prime}(2) $$

Step-by-Step Solution

Answer

The value of the second derivative \(g''(2)\) is 260

Step 1: Differentiate the function \(g(t)\) once

To find the first derivative of \(g(t)=5 t^{4}+10 t^{2}+3\), use the power rule for differentiation, which states that the derivative of \(t^n\) with respect to \(t\) is \(n*t^{n-1}\). The first derivative \(g'(t)\) is found by differentiating each term of \(g(t)\) with respect to \(t\). This results in: \(g'(t)= 20t^{3} + 20t\).

Step 2: Differentiate the function \(g'(t)\) again

The second derivative \(g''(t)\) is found by differentiating the first derivative \(g'(t)\). Again, use the power rule for each term in \(g'(t)\): \(g''(t) = 60t^{2} + 20\).

Step 3: Evaluate the second derivative at \(t=2\)

Finally, evaluate the second derivative \(g''(t)\) at \(t=2\). Substitute \(2\) into the function \(g''(t)\): \(g''(2) = 60*(2)^2 + 20 = 260\).

Key Concepts

Power Rule

The power rule is a basic principle in differentiation that simplifies the process of finding derivatives, especially for polynomial functions. It specifically deals with terms that are in the format of a variable raised to a power, like \( t^n \).
To apply the power rule, you take the exponent of the term and multiply it by the coefficient, then subtract one from the exponent. For instance, if you differentiate \( t^4 \), you multiply the 4 by the coefficient (if there's no number, it means the coefficient is 1) to get 4, and then you reduce the power by 1, resulting in \( 4t^3 \).

• The rule can be concisely stated as: if \( f(t) = at^n \), then the derivative \( f'(t) = nat^{n-1} \).
• This makes finding the derivatives of polynomial terms straightforward and fast.

Importantly, you apply the power rule separately to each term in a function and then sum them up for the total derivative. This method is straightforward and incredibly useful for differentiation in calculus.

Differentiation

Differentiation is a key operation in calculus used to determine the derivative of a function. The derivative measures how a function's value changes as its input changes.
Think of it as finding the slope of the function at any point. This allows us to understand the rate of change in a variety of real-world scenarios.

• To differentiate a function, each term has to undergo differentiation individually.
• Each term's derivative shows how that specific part of the function behaves or changes the function's overall outcome.

Differentiating multiple times grants access to higher-order derivatives. For example, the second derivative, denoted as \( g''(t) \), provides insight into the curvature or concavity of the function's graph, an important aspect in determining points of inflection and other graph characteristics.

Calculus

Calculus is a branch of mathematics focused on change and motion. Its two main components are differentiation (which finds rates of change) and integration (which finds quantities based on rates of change).
Derivatives (from differentiation) allow us to model dynamic situations where change is continuous, helping to resolve complex real-world problems.

• Calculus plays a vital role in fields such as physics, engineering, economics, and beyond.
• Its concepts are applied in optimizing functions, finding speeds and accelerations, and understanding scientific phenomena.

Practically, learning calculus strengthens problem-solving skills and broadens analytical thinking, equipping students to handle challenging tasks in a structured way. Understanding foundational concepts like the power rule in differentiation aids significantly in grasping more intricate calculus principles.