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find the indicated derivatives. $$\frac{d r}{d s} \text { if } r=s^{3}-2 s^{2}+3$$

Short Answer

Expert verified
\( \frac{dr}{ds} = 3s^2 - 4s \)

Step by step solution

01

Identify the Function

We are given the function \( r(s) = s^3 - 2s^2 + 3 \). Our task is to find the derivative of \( r \) with respect to \( s \), that is \( \frac{dr}{ds} \).
02

Apply the Power Rule

The power rule states that \( \frac{d}{ds}[s^n] = ns^{n-1} \). We apply this rule to each term of the function \( r \).
03

Differentiate Each Term

Differentiate each term of the function separately:- For \( s^3 \), the derivative is \( 3s^2 \).- For \( -2s^2 \), the derivative is \( -4s \).- The constant 3 has a derivative of 0.
04

Combine the Derivatives

Combine the derivatives calculated in the previous step to obtain the derivative of the entire function:\[ \frac{dr}{ds} = 3s^2 - 4s + 0 \]. Thus, simplify to get \( \frac{dr}{ds} = 3s^2 - 4s \).

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Key Concepts

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

Power Rule
The power rule is a fundamental tool in calculus for finding derivatives. It's a simple yet powerful technique used for turning functions into their rate of change. In essence, if you have a function of the form \( s^n \), the power rule states that its derivative is \( ns^{n-1} \). This means you multiply the power by the coefficient and then reduce the power by one.

For example, consider \( s^3 \). Using the power rule, the derivative is simply \( 3s^2 \) because you bring down the exponent 3 as a coefficient and subtract one from the exponent. Similarly, for \( -2s^2 \), the derivative becomes \( -4s \). This is because you multiply the coefficient -2 by the power 2, and subtract one from the power, resulting in the new power of 1.
  • The Power Rule is helpful for differentiating polynomial terms.
  • It simplifies the process of finding derivatives of polynomials considerably.
This rule is often among the first differentiation techniques taught because of its straightforward application and utility across various problems.
Differentiation
Differentiation is a process in calculus used to compute the derivative of a function. The derivative provides crucial information about the function's rate of change or slope. When differentiating, you essentially find how a function's output changes with respect to changes in its input. This change is visualized as the slope of the function's graph at any given point.

The primary purpose of differentiation is to determine the instantaneous rate of change of a quantity. For instance, differentiating \( r(s) = s^3 - 2s^2 + 3\) with respect to \( s \) involves applying the power rule to each term:
  • For \( s^3 \), the derivative is \( 3s^2 \).
  • For \( -2s^2 \), the derivative is \( -4s \).
  • A constant term like 3 has a derivative of 0.
By combining these derivatives, we get the overall derivative of the function. In this example, the derivative \( \frac{dr}{ds} \) is simply \( 3s^2 - 4s \), showing the changes for each value of \( s \). Differentiation is indispensable in analyzing dynamic systems and solving real-world problems.
Calculus
Calculus is a branch of mathematics that studies change and motion. It is divided into two primary areas: differentiation and integration. While differentiation focuses on finding rates of change, integration deals with accumulating quantities, like areas under curves.

In the context of differentiation, calculus provides a structured method for finding derivatives of functions. This approach helps solve complex mathematical problems and model various physical phenomena. For example, with the function \( r(s) = s^3 - 2s^2 + 3\), calculus allows us to smoothly apply the power rule to derive \( \frac{dr}{ds} = 3s^2 - 4s \).

Some key features of calculus include:
  • Its ability to handle smooth and continuous functions, enabling the exploration of precise physical concepts.
  • Utility in various fields such as physics, engineering, economics, and any domain involving continuous change.
Calculus serves as a foundational tool in science and engineering, opening doors to advances in technology and our understanding of the universe.

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Most popular questions from this chapter

In Exercises \(35-40,\) write a differential formula that estimates the given change in volume or surface area. $$ \begin{array}{l}{\text { The change in the surface area } S=6 x^{2} \text { of a cube when the edge }} \\ {\text { lengths change from } x_{0} \text { to } x_{0}+d x}\end{array} $$

Use a CAS to perform the following steps. \begin{equation} \begin{array}{l}{\text { a. Plot the equation with the implicit plotter of a CAS. Check to }} \\ {\text { see that the given point } P \text { satisfies the equation. }} \\ {\text { b. Using implicit differentiation, find a formula for the deriva- }} \\ {\text { tive } d y / d x \text { and evaluate it at the given point } P .}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { c. Use the slope found in part (b) to find an equation for the tan- }} \\ {\text { gent line to the curve at } P \text { . Then plot the implicit curve and }} \\ {\text { tangent line together on a single graph. }}\end{array} \end{equation} \begin{equation} x^{5}+y^{3} x+y x^{2}+y^{4}=4, \quad P(1,1) \end{equation}

Use a CAS to perform the following steps. \begin{equation} \begin{array}{l}{\text { a. Plot the equation with the implicit plotter of a CAS. Check to }} \\ {\text { see that the given point } P \text { satisfies the equation. }} \\ {\text { b. Using implicit differentiation, find a formula for the deriva- }} \\ {\text { tive } d y / d x \text { and evaluate it at the given point } P .}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { c. Use the slope found in part (b) to find an equation for the tan- }} \\ {\text { gent line to the curve at } P \text { . Then plot the implicit curve and }} \\ {\text { tangent line together on a single graph. }}\end{array} \end{equation} \begin{equation} y^{3}+\cos x y=x^{2}, \quad P(1,0) \end{equation}

Estimate the allowable percentage error in measuring the diameter \(D\) of a sphere if the volume is to be calculated correctly to within 3\(\% .\)

Oil spill An explosion at an oil rig located in gulf waters causes an elliptical oil slick to spread on the surface from the rig. The slick is a constant 9 in. thick. After several days, when the major axis of the slick is 2 mi long and the minor axis is 3\(/ 4\) mi wide, it is determined that its length is increasing at the rate of 30 \(\mathrm{ft} / \mathrm{hr}\) , and its width is increasing at the rate of 10 \(\mathrm{ft} / \mathrm{hr}\) . At what rate (in cubic feet per hour) is oil flowing from the site of the rig at that time?

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