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Chapter 4: Problem 54

Find formulas for the functions described. A function of the form \(y=a \sin \left(b t^{2}\right)\) whose first critical point for positive \(t\) occurs at \(t=1\) and whose derivative is 3 when \(t=2.\)

Short Answer

Expert verified
The function is \( y = \frac{3}{2\pi} \sin\left(\frac{\pi}{2} t^2\right) \).

Step by step solution

01

Understanding the Critical Point

For the first critical point of the function to occur at \(t = 1\), the derivative \(y'\) must be equal to zero at this point. This means we need to compute \(y'\) and set it to zero to find the relationship between the constants \(a\) and \(b\).
02

Derive the Function

Start with the given function, \(y = a \sin(bt^2)\). To find the derivative, use the chain rule: \(y' = a \cos(bt^2) \cdot \frac{d}{dt}(bt^2) = a \cdot \cos(bt^2) \cdot 2bt\).
03

Set Derivative to Zero for Critical Point

To find when the first critical point occurs at \(t=1\), set the derivative equal to zero: \[ y' = a \cdot \cos(b) \cdot 2b \cdot 1 = 0. \] This implies \(\cos(b)=0\). Hence, \(b = \frac{pi}{2}\) for some odd integer \(n\). Choosing \(n=1\), we have \(b = \frac{\pi}{2}\).
04

Use Derivative at t=2 for a Second Equation

The derivative is given as 3 when \(t=2\). Substitute \(t=2\) in the derivative equation: \[ y' = a \cdot \cos\left(\frac{\pi}{2} \cdot 4 \right) \cdot 2 \cdot \frac{\pi}{2} \cdot 2. \] Since \(\cos(2\pi) = 1\), \[ 3 = a \cdot 1 \cdot \frac{4\pi}{2} = 2a\pi. \] Solving for \(a\) gives \(a=\frac{3}{2\pi}\).
05

Formulate the Function

Now that we have both \(a\) and \(b\), substitute them back into the original function: \[ y = \frac{3}{2\pi} \sin\left(\frac{\pi}{2} t^2\right). \] This is the formula for the function.

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Key Concepts

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

Chain rule
When we need to find the derivative of a function composed of other functions, we apply the chain rule. Specifically, it helps to differentiate a composite function by taking the derivative of the outer function and multiplying it by the derivative of the inner function.For the function given in the exercise, let's break it down:
- Original function: \[ y = a \sin(bt^2) \]- Outer function: \(\sin(x)\) where \(x = bt^2\)- Inner function: \(x = bt^2\)Following the chain rule:1. Determine the derivative of the outer function: - The derivative of \(\sin(x)\) is \(\cos(x)\). - So, we write it as \(\cos(bt^2)\).2. Then, determine the derivative of the inner function: - The derivative of \(bt^2\) is \(2bt\).3. Finally, multiply them together: - The derivative of the entire expression becomes \(a \cdot \cos(bt^2) \cdot 2bt\).The chain rule is crucial because it easily handles function layers often seen with trigonometric and polynomial combinations.
Trigonometric functions
Trigonometric functions, such as \(\sin\), \(\cos\), and \(\tan\), are vital in calculus, as they describe periodic phenomena. In this exercise, we focus on the sine function, \(\sin(x)\), and its properties.To understand how critical points arise, particularly how the derivative's zero value signifies these, consider:
  • The function \(y = a\sin(bt^2)\) features a trigonometric function within it, controlling the oscillation based on \(t\).
  • Critical points occur when the derivative equals zero. For \(\cos(x)\) (as derived), this happens when its argument, here \(bt^2\), equals angles like \(\pi/2\) that make \(\cos(x) = 0\).
The insight that \(b = \pi/2\) by choosing an odd multiple supports such a critical point scenario.Considering trigonometric identities and values is essential to solving equations involving derivatives.
Derivatives
The derivative of a function provides critical insights into its behavior, specifically:
  • Momentary rate of change: How fast or slow the function is changing at a certain point.
  • Identifying critical points: Where the function "flattens," indicating potential maxima, minima, or points of inflection, occurs where the derivative equals zero.
In the problem, the computed derivative \( y' = a \cdot \cos(bt^2) \cdot 2bt \) leads to two scenarios:1. At \(t = 1\), we set the derivative to zero: - This determined \( b \) value to achieve the sine function's critical point.2. At \(t = 2\), knowing the derivative equates to 3: - This information allowed us to solve for \( a \), necessary for controlling the overall function's rate of change.Thus, by solving equations with derivatives, we tuned the parameters \(a\) and \(b\) to achieve specific function characteristics on specified criteria.

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Most popular questions from this chapter

At time \(t\), the position of a particle moving on a curve is given by \(x(t)=3 t^{2}-1\) and \(y(t)=t^{2}-3 t .\) At \(t=2;\) (a) What is the position of the particle? (b) What is the slope of the curve? (c) What is the speed of the particle?

A particle moves in the \(x y\) -plane with position at time \(t\) given by \(x=\sin t\) and \(y=\cos (2 t)\) for \(0 \leq t<2 \pi.\) (a) At what time does the particle first touch the \(x\) axis? What is the speed of the particle at that time? (b) Is the particle ever at rest? (c) Discuss the concavity of the graph.

In many applications, we want to maximize or minimize some quantity subject to a condition. Such constrained optimization problems are solved using Lagrange multipliers in multivariable calculus;show an alternate method.With quantities \(x\) and \(y\) of two raw materials available, \(Q=x^{1 / 2} y^{1 / 2}\) thousand items can be produced at a cost of \(C=2 x+y\) thousand dollars. Using the following steps, find the minimum cost to produce 1 thousand items. (a) Graph \(x^{1 / 2} y^{1 / 2}=1 .\) On the same axes, graph \(2 x+y=2,2 x+y=3,\) and \(2 x+y=4\) (b) Explain why the minimum cost occurs at a point at which a cost line is tangent to the production curve \(Q=1\) (c) Using your answer to part (b) and implicit differentiation to find the slope of the curve, find the minimum cost to meet this production level.

Determine whether the limit exists, and where possible evaluate it. \(\lim x^{a} \ln x,\) where \(a\) is \(x \rightarrow 0+\) a positive constant.

Evaluate the limits where $$f(t)=\left(\frac{3^{t}+5^{t}}{2}\right)^{1 / t} \quad \text { for } t \neq 0$$ $$\lim _{t \rightarrow-\infty} f(t)$$
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