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Chapter 3: Problem 12

Find \(D_{t}{ }^{3} s\) if \(s=\sqrt{4 t+1}\).

Short Answer

Expert verified
The third derivative is \( 24 (4t + 1)^{-5/2} \).

Step by step solution

01

Identify the given function

The problem gives the function as: \[ s = \sqrt{4t + 1} \]
02

Compute the first derivative

To find the first derivative of \( s \) with respect to \( t \), use the chain rule. \[ \frac{d}{dt} (\sqrt{4t + 1}) = \frac{1}{2} ( 4t + 1 )^{-1/2} \times 4 \] Which simplifies to: \[ \frac{2}{\sqrt{4t + 1}} \]
03

Compute the second derivative

To find the second derivative, apply the derivative operator to the first derivative. \[ \frac{d}{dt} \frac{2}{\sqrt{4t + 1}} \] Using the chain rule again, this becomes: \[ 2 \times (4t + 1)^{-3/2} \times (-2) \] Which simplifies to: \[ -4 (4t + 1)^{-3/2} \]
04

Compute the third derivative

To find the third derivative, apply the derivative operator to the second derivative. \[ \frac{d}{dt} [-4 (4t + 1)^{-3/2}] \] Using the chain rule, this becomes: \[ -4 \times (-3/2) \times (4t + 1)^{-5/2} \times 4 \] Which simplifies to: \[ 24 (4t + 1)^{-5/2} \]

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Key Concepts

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

chain rule
The chain rule is a fundamental technique in calculus. It helps when you need to differentiate a composite function. Essentially, it states that if you have a function composed of two functions, say \(h(x) = f(g(x))\), then its derivative is formed by multiplying the derivative of the outer function by the derivative of the inner function. For a clearer understanding, let鈥檚 take \(s = \sqrt{4t+1}\). First, we set \(u = 4t + 1\). Thus, our function becomes \(s = \sqrt{u}\). The chain rule tells us:
\[ \frac{ds}{dt} = \frac{ds}{du} \times \frac{du}{dt} \]
This step-by-step approach can significantly simplify the differentiation process.
first derivative
The first derivative helps you understand how a function changes at any given point. For the function \(s = \sqrt{4t + 1}\), we apply the chain rule. First, let鈥檚 rewrite \(s\) using an exponent: \(s = (4t + 1)^{1/2}\).
Using the chain rule, the first derivative becomes:
\[ \frac{d}{dt} (4t + 1)^{1/2} = \frac{1}{2} (4t + 1)^{-1/2} \times 4 \]
Simplifying this gives:
\[ \frac{2}{\sqrt{4t + 1}} \]
This first derivative tells us the rate of change of \(s\) concerning \(t\).
second derivative
The second derivative gives insight into the concavity or curvature of a function. For \(s = \sqrt{4t + 1}\), we already found the first derivative: \( \frac{2}{\sqrt{4t + 1}} \). To find the second derivative, we again use the chain rule:
\[ \frac{d}{dt} \frac{2}{\sqrt{4t + 1}} \]
This becomes:
\[ 2 \times (4t + 1)^{-3/2} \times (-2) \]
Simplifying, we get:
\[ -4 (4t + 1)^{-3/2} \]
This second derivative indicates how the rate of change is itself changing, providing further depth in understanding the behavior of \(s\).
third derivative
The third derivative informs us about the rate of change of the second derivative, offering insights into the jerk or how the acceleration changes over time. For \(s = \sqrt{4t + 1}\), we need to differentiate the second derivative:
\[ \frac{d}{dt} [-4 (4t + 1)^{-3/2}] \]
Applying the chain rule for the third time, we get:
\[ -4 \times (-3/2) \times (4t + 1)^{-5/2} \times 4 \]
Which simplifies to:
\[ 24 (4t + 1)^{-5/2} \]
This final derivative shows the nuanced changes in the function's rate of change at an even deeper level, adding another layer to our understanding of \(s\).

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Most popular questions from this chapter

Find the slope of the tangent line to the graph at the point \(\left(x_{1}, y_{1}\right) .\) Make a table of values of \(x, y\), and \(m\) at various points on the graph, and include in the table all points where the graph has a horizontal tangent. Draw a sketch of the graph. $$ y=4 x^{3}-13 x^{2}+4 x-3 $$

The motion of a particle is along a horizontal line according to the given equation of motion, where \(s \mathrm{ft}\) is the directed distance of the particle from a point \(O\) at \(t \mathrm{sec}\). The positive direction is to the right. Determine the intervals of time when the particle is moving to the right and when it is moving to the left. Also determine when the particle reverses its direction. Show the behavior of the motion by a figure similar to Fig. 3.2.2, choosing values of \(t\) at random but including the values of \(t\) when the particle reverses its direction. $$ s=2 t^{3}-3 t^{2}-12 t+8 $$

Find the derivative of the given function. $$ H(z)=\left(z^{3}-3 z^{2}+1\right)^{-3} $$

In Exercises 22 and 23, a particle is moving along a straight line according to the given equation of motion, where \(s \mathrm{ft}\) is the directed distance of the particle from the origin at \(t \mathrm{sec}\). If \(v \mathrm{ft} / \mathrm{sec}\) is the velocity and \(a \mathrm{ft} / \mathrm{sec}^{2}\) is the acceleration of the particle at \(t \mathrm{sec}\), find \(v\) and \(a\) in terms of \(t\). Also find when the acceleration is zero and the intervals of time when the particle is moving toward the origin and when it is moving away from the origin. $$ s=\frac{1}{4} t^{4}-2 t^{3}+4 t^{2} $$

In Exercises 1 through 8 , find the slope of the tangent line to the graph at the point \(\left(x_{1}, y_{1}\right) .\) Make a table of values of \(x, y\), and \(m\) at various points on the graph, and include in the table all points where the graph has a horizontal tangent. Draw a sketch of the graph. $$ y=9-x^{2} $$
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