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Magazine Chapter 27: Problem 6
Find the derivatives of the given functions. $$s=5 \sin (7-3 t)$$
Short Answer
Expert verified
The derivative is \( s'(t) = -15 \cos(7 - 3t) \).
Step by step solution
01
- Identify the Function Type
The function given is \( s = 5 \sin(7 - 3t) \). This is a trigonometric function, specifically a sine function, where the argument of the sine function is \( 7 - 3t \).
02
- Apply the Chain Rule
To find the derivative of \( s = 5 \sin(7 - 3t) \), we need to use the chain rule. The chain rule states that if you have a composite function \( \sin(g(t)) \), its derivative is \( \cos(g(t)) \cdot g'(t) \). Here, \( g(t) = 7 - 3t \).
03
- Differentiate the Outside Function
Differentiate the sine function with respect to its argument \( 7-3t \). The derivative of \( \sin(u) \) with respect to \( u \) is \( \cos(u) \). Therefore, the derivative of \( \sin(7 - 3t) \) with respect to \( 7 - 3t \) is \( \cos(7 - 3t) \).
04
- Differentiate the Inside Function
Differentiate \( 7 - 3t \) with respect to \( t \). The derivative \( \frac{d}{dt}(7 - 3t) = -3 \) since the derivative of \( 7 \) is \( 0 \) and the derivative of \( -3t \) is \( -3 \).
05
- Combine Using Chain Rule
Combine the results using the chain rule. Multiply the derivative of the outside function by the derivative of the inside function: \( 5 \cdot \cos(7 - 3t) \cdot (-3) \).
06
- Simplify the Expression
Simplify the expression to find the derivative of the function. The derivative \( s'(t) = 5 \times (-3) \times \cos(7-3t) = -15 \cos(7-3t) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Understanding Derivatives
Derivatives are a fundamental concept in calculus. They represent the rate at which a function changes at any given point.
- Consider a function that describes a car's position over time. The derivative of this function would tell you the car's speed at any specific moment.
- For the function in the exercise, which is a trigonometric function involving sine, the derivative will tell us how rapidly the sine wave is changing as the variable \( t \) changes.
- In our specific problem, we are dealing with the derivative of \( s = 5 \sin(7 - 3t) \) with respect to \( t \).
Applying the Chain Rule
The chain rule is a powerful technique in calculus used for differentiating composite functions. A composite function is one where one function is nested within another, much like layers of an onion.
- The outer function here is \( \, \sin \, \) and its derivative is \( \, \cos \, \).
- For example, if you have a function \( f(g(x)) \), where \( g(x) \) is inside \( f(x) \), you're dealing with a composite function.
- The chain rule helps us differentiate such layers by providing a method to take the derivative of the outer function and multiply it by the derivative of the inner function.
- The outer function here is \( \, \sin \, \) and its derivative is \( \, \cos \, \).
- Thus, the derivative of \( \sin(u) \) is \( \cos(u) \) where \( u = 7 - 3t \).
- Next, the inner function \( 7 - 3t \) must be differentiated with respect to \( t \), which results in \( -3 \).
Trigonometric Functions in Calculus
Trigonometric functions like \( \sin \, \) and \( \cos \, \) are a cornerstone of calculus due to their periodic properties.
Trigonometric derivatives like these help model and analyze a wide array of physical and theoretical systems, making them both challenging and profoundly useful in calculus.
- These functions often appear in problems related to waves, oscillations, and circular motion because of their cyclical nature, which aptly models such phenomena.
- When differentiating trigonometric functions, it's vital to recall their specific derivatives: the derivative of \( \sin(u) \) is \( \cos(u) \), and vice-versa, the derivative of \( \cos(u) \) is \( -\sin(u) \).
Trigonometric derivatives like these help model and analyze a wide array of physical and theoretical systems, making them both challenging and profoundly useful in calculus.
