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Find the derivative of the function. \(f(r)=\pi r^{2}+2 \pi r\)

Short Answer

Expert verified
The derivative of the function \(f(r) = \pi r^2 + 2\pi r\) is \(f'(r) = 2\pi r + 2\pi\).

Step by step solution

01

Identify the terms of the function

In this function, we have two terms: \(\pi r^2\) and \(2\pi r\). We will find the derivative of each term separately and then add them together to find the derivative of the entire function.
02

Differentiate the first term

The first term in our function is \(\pi r^2\). Using the power rule, the derivative of \(r^n\) with respect to r is \(nr^{n-1}\). Therefore, the derivative of \(\pi r^2\) with respect to r is: \(2\pi r^{2-1} = 2\pi r\)
03

Differentiate the second term

The second term in our function is \(2\pi r\). Using the power rule, the derivative of \(r^1\) with respect to r is simply 1. Therefore, the derivative of \(2\pi r\) with respect to r is: \((2\pi) * 1 = 2\pi\)
04

Combine the derivatives of each term

Now that we have found the derivatives of both terms in our function, we can add them together to find the derivative of the entire function: \(f'(r) = 2\pi r + 2\pi\) So, the derivative of the function \(f(r) = \pi r^2 + 2\pi r\) is: \(f'(r) = 2\pi r + 2\pi\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Differentiation
When we talk about function differentiation in calculus, we're essentially discussing how we can determine the rate at which a function is changing at any point. This process involves finding what's known as the "derivative" of the function. Differentiation is a fundamental concept in calculus and is used to analyze the behavior of functions. By identifying how a function changes in response to a change in its input, we gain insights that are valuable for various fields like physics, engineering, and economics.
  • A derivative tells us the slope of the function at a particular point.
  • It provides information about the rate of change of the function's value with respect to its input variable.
  • This rate of change is crucial for understanding dynamic systems in real-world scenarios.
Let's take the example of the function we've differentiated: \( f(r) = \pi r^2 + 2 \pi r \). Differentiating it gives us \( f'(r) = 2\pi r + 2\pi \), which shows how the function changes as \( r \) varies.
Power Rule
The power rule is a crucial technique in calculus that simplifies the differentiation process, especially for polynomial functions. It states that if you have a term in the form \( x^n \), where \( n \) is any real number, the derivative is \( nx^{n-1} \). This rule makes it easy to differentiate polynomial terms by simply multiplying the power of \( x \) by the coefficient, and then subtracting one from the power. In our function \( f(r) = \pi r^2 + 2 \pi r \), the power rule is used to differentiate each term:
  • For the term \( \pi r^2 \), applying the power rule gives us \( 2\pi r^{2-1} = 2\pi r \).
  • For the term \( 2\pi r \), which is equivalent to \( 2\pi r^1 \), the power rule gives \( (1)(2\pi)r^{1-1} = 2\pi \).
The power rule thus helps us efficiently find the derivative, making it one of the first techniques students learn when tackling differential calculus.
Calculus
Calculus, a branch of mathematics, is fundamentally about change. It provides tools for modeling and analyzing dynamic systems and is composed of two main areas: differentiation and integration. Differentiation, as we've discussed, deals with finding the rate at which a function is changing.
  • The derivative, a central element in calculus, offers insights into the instantaneous rate of change, akin to velocity (change of position over time) in physics.
  • Calculus is widely used in various fields such as physics for motion analysis, biology for population growth models, and economics for optimizing costs and revenues.
One of the reasons calculus is so powerful is because it can handle curves and non-linear relationships which are prevalent in real-world scenarios. Regarding our function \( f(r) = \pi r^2 + 2 \pi r \), calculus allows us to derive the function's behavior through differentiation, revealing critical points like maxima, minima, and inflection points which are essential for understanding the geometry and dynamics of the problem.

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Most popular questions from this chapter

Electric Potential Suppose that a ring-shaped conductor of radius \(a\) carries a total charge \(Q\). Then the electrical potential at the point \(P\), a distance \(x\) from the center and along the line perpendicular to the plane of the ring through its center, is given by $$ V(x)=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{\sqrt{x^{2}+a^{2}}} $$ where \(\varepsilon_{0}\) is a constant called the permittivity of free space. The magnitude of the electric field induced by the charge at the point \(P\) is \(E=-d V / d x\), and the direction of the field is along the \(x\) -axis. Find \(E\).

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If a right circular cylinder of radius \(a\) is filled with water and rotated about its vertical axis with a constant angular velocity \(\omega\), then the water surface assumes a shape whose cross section in a plane containing the vertical axis is a parabola. If we choose the \(x y\) -system so that the \(y\) -axis is the axis of rotation and the vertex of the parabola passes through the origin of the coordinate system, then the equation of the parabola is $$y=\frac{\omega^{2} x^{2}}{2 g}$$ where \(g\) is the acceleration due to gravity. Find the angle \(\alpha\) that the tangent line to the water level makes with the \(x\) -axis at any point on the water level. What happens to \(\alpha\) as \(\omega\) increases? Interpret your result.

Find an equation of the tangent line to the given curve at the indicated point. $$ x^{2} y^{2}=(y+1)^{2}\left(4-y^{2}\right) ; \quad(-2 \sqrt{3}, 1) $$

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