Question

Use a graphing calculator or a CAS If a point moves along a line so that its distance \(s\) (in feet) from 0 is given by \(s=t+t \cos ^{2} t\) at time \(t\) seconds, find its instantaneous velocity at \(t=3\).

Short answer

The instantaneous velocity at \(t=3\) is approximately \(2.82\) ft/s.

Step-by-step solution

Understand the Problem

The problem asks for the instantaneous velocity of a point moving along a line, given by the distance function \(s(t) = t + t \cos^2 t\), at \(t = 3\) seconds. To find instantaneous velocity, we need the derivative of the distance function, \(s(t)\), with respect to time \(t\).

Differentiate the Function

First, identify the terms in the distance function: \(s(t) = t + t \cos^2 t\). The function consists of two terms: \(t\) and \(t \cos^2 t\). Differentiate each term with respect to \(t\).1. The derivative of \(t\) is 1.2. Use the product rule to differentiate \(t \cos^2 t\): - Let \(u = t\) and \(v = \cos^2 t\). - \( u' = 1\) and \(v' = 2 \cos t (-\sin t) = -2 \cos t \sin t = -\sin(2t)\). Product Rule: \( (uv)' = u'v + uv' \to (t \cos^2 t)' = (1)(\cos^2 t) + t(-\sin(2t))\). So, \( (t \cos^2 t)' = \cos^2 t - t \sin(2t)\).Combine the derivatives: \( s'(t) = 1 + \cos^2 t - t \sin(2t) \).

Substitute \(t = 3\) into the Derivative

Substitute \(t = 3\) into the derivative equation to find the instantaneous velocity:\[ s'(3) = 1 + \cos^2(3) - 3 \sin(2 \times 3) \].

Calculate the Values

Calculate \(\cos^2(3)\) and \(-3 \sin(6)\):- Remember to check if the calculator is in radians mode if you're using one.- \(\cos(3) \approx -0.9899924966\), thus \(\cos^2(3) = (-0.9899924966)^2 \approx 0.9800665778\).- \(\sin(6) \approx -0.2794154982\), thus \(-3 \sin(6) = -3(-0.2794154982) = 0.8382464946\).Then substitute these values back into the equation:\[ s'(3) = 1 + 0.9800665778 + 0.8382464946 \approx 2.8183130724 \].

Conclusion

The instantaneous velocity of the point at \(t = 3\) seconds is approximately \(2.82\) feet per second.

Key concepts

Calculus

Calculus is a branch of mathematics that deals with rates of change and accumulation. It is a fundamental tool for analyzing problems involving motion and change, just like in our exercise which involves finding instantaneous velocity.

In calculus, two primary concepts stand out:

• Differentiation: This process helps us find the rate at which a quantity changes. In our context, it's used to find instantaneous velocity, which is the derivative of the distance function.
• Integration: This is the process of calculating the accumulated quantity, such as areas under curves or the total distance covered over time.

To solve problems like ours, understanding the basics of calculus, particularly differentiation, is crucial. This allows us to analyze how a point's position changes with respect to time, capturing dynamic behavior accurately.

Differentiation

Differentiation is a way of finding the exact rate at which one quantity changes compared to another. It helps to calculate the derivative of functions, which indicates the function's slope at any given point.

To differentiate a function, follow these general steps:

• Understand the Function: Identify the parts of the function you need to differentiate.
• Apply Differentiation Rules: Use rules like the product rule or chain rule to calculate derivatives of complex functions.
• Simplify the Result: Reduce the differentiated equation to its simplest form.

In our exercise, the key lies in finding the rate of change of distance with respect to time. This is achieved by differentiating the distance function to obtain the velocity function, which provides the instantaneous velocity at any given time.

Product Rule

The product rule in calculus is particularly useful when differentiating products of two functions. This situation arises often in physics and engineering where a quantity is a product of two varying factors.

The product rule states that if you have two functions, say \( u(t) \) and \( v(t) \), then the derivative of their product, \( uv \), is:\[(uv)' = u'v + uv'\]
In our problem, the distance is defined as a product of \( t \) and \( \cos^2 t \). So, utilizing the product rule becomes critical.

By assigning:

• \( u = t \) with \( u' = 1 \)
• \( v = \cos^2 t \)

We find \( v' \) using the chain rule resulting in \( -\sin(2t) \), and then apply the product rule to find the derivative correctly. This step ensures that we capture all variations accurately in the velocity function.

Trigonometric Functions

Trigonometric functions like sine and cosine are fundamental in calculus, especially in problems involving cycles or periodic motion. These functions describe the relationships between the angles and sides of triangles and are used to model wave-like patterns in various phenomena.

In our exercise, \( \cos^2 t \) is a trigonometric function embedded within the distance equation. Differentiating this requires understanding how trigonometric derivatives work:

• The derivative of \( \cos t \) is \( -\sin t \).
• The derivative of \( \sin t \) is \( \cos t \).
• The derivative of \( \cos^2 t \), via the chain rule, results in \( -\sin(2t) \).

These derivatives help us compute the equation's rate of change, crucial for finding the instantaneous velocity of the point as it moves along its path. Understanding these concepts allows us to precisely and confidently tackle problems involving trigonometric functions in calculus.