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Chapter 3: Problem 10

In Exercises \(9-12,\) an object moves along the \(x\) -axis so that its position at any time \(t \geq 0\) is given by \(x(t)=s(t) .\) Find the velocity of the object as a function of \(t .\) $$s=t \cos (\pi-4 t)$$

Short Answer

Expert verified
The velocity of the object as a function of time \( t \) is given by \( v(t) = \cos(\pi - 4t) - t \sin(\pi - 4t) * (\pi - 4) \)

Step by step solution

01

Identify the function to differentiate

Given the position function \( s(t) = t \cos(\pi - 4t) \) which represents the position of the object at time \( t \geq 0 \). According to physics, the velocity of an object which moves on a straight path is the derivative of the position function with respect to time. Therefore, to find the velocity function, compute the derivative of the position function \( s(t) \) using the product rule and chain rule of differentiation.
02

Apply the product and chain rule

Recall the product rule states, if \( u(t) \) and \( v(t) \) are functions of \( t \) then the derivative of their product is \( u'(t)v(t) + u(t)v'(t) \). The chain rule states that the derivative of a composite function \( f(g(x)) \) is \( f'(g(x)) * g'(x) \). Here, let \( u(t) = t \) and \( v(t) = \cos(\pi - 4t) \). So, \( u'(t) = 1 \) and using the chain rule \( v'(t) = - \sin(\pi - 4t) * (\pi - 4) \). Inserting these into the product rule formula, \( s'(t) = u'(t)v(t) + u(t)v'(t) = 1*\cos(\pi - 4t) + t * - \sin(\pi - 4t) * (\pi - 4) = \cos(\pi - 4t) - t \sin(\pi - 4t) * (\pi - 4) \), this is the velocity function.
03

Present the final answer

The velocity of the object as a function of time \( t \) is given by \( v(t) = \cos(\pi - 4t) - t \sin(\pi - 4t) * (\pi - 4) \)

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Key Concepts

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

Understanding the Product Rule
When you're trying to find the derivative of a product of two functions, you use the product rule. This rule is a fundamental part of calculus, making it easier to tackle more complex problems. The product rule states that the derivative of a product of two functions, \( u(t) \) and \( v(t) \), is
  • \( u'(t)v(t) + u(t)v'(t) \)
This means you first differentiate \( u(t) \), keeping \( v(t) \) the same, and then differentiate \( v(t) \) while keeping \( u(t) \) the same. You combine these two results to get your answer. For example, if \( u(t) = t \) and \( v(t) = \cos(\pi - 4t) \), the product rule helps us find the derivative of this product. Apply the rule by first differentiating \( t \) to get \( 1 \) and keeping \( \cos(\pi - 4t) \) the same, and vice versa.
Diving into the Chain Rule
The chain rule is essential when dealing with composite functions. It helps us understand how to find the derivative of a function within another function. When you have a composite function \( f(g(t)) \), the chain rule says that the derivative is
  • \( f'(g(t)) \cdot g'(t) \).
This means you first find the derivative of the outer function, keeping the inner function the same, and multiply by the derivative of the inner function. In the example given, to differentiate \( \cos(\pi - 4t) \), identify it as a function within another and use the chain rule. Keep \( \cos \) and find its derivative as \( -\sin \). Now multiply it by the derivative of the inner function \( \pi - 4t \), which is \( -4 \). Adjustments like these make the chain rule very powerful for solving complex derivatives.
Grasping the Basics of Derivatives
Derivatives are central to calculus and represent how a function changes at any given point. When we talk about the velocity function of an object, we're really talking about the derivative of its position function.
  • The derivative tells you the rate of change or slope.
  • In physics, it helps understand how velocity changes over time.
For the position function \( s(t) = t \cos(\pi - 4t) \), the derivative gives us the velocity function. To compute this, you break it down using the product rule and the chain rule, as discussed earlier. By using these derivatives, you can determine how the position function changes with respect to time, offering insight into the object's speed and direction. Remember, understanding derivatives is crucial for grasping concepts related to motion and rate of change.

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