Find the rate of change of [tex]y[/tex] with respect to [tex]x[/tex] at the indicated value of [tex]x[/tex].
[tex]y=\csc x-2 \cos x ; \quad x=\frac{\pi}{6}[/tex]
Answer (1)
Find the derivative of Y = csc x − 2 cos x with respect to x , which is d x d Y = − csc x cot x + 2 sin x .
Evaluate the derivative at x = 6 π using the known values of csc ( 6 π ) = 2 , cot ( 6 π ) = 3 , and sin ( 6 π ) = 2 1 .
Substitute these values into the derivative expression: − ( 2 ) ( 3 ) + 2 ( 2 1 ) = − 2 3 + 1 .
The rate of change of Y with respect to X at x = 6 π is 1 − 2 3 .
Explanation
Problem Analysis We are given the function Y = csc x − 2 cos x and we need to find the rate of change of Y with respect to X at x = 6 π . This means we need to find the derivative d X d Y and evaluate it at x = 6 π .
Finding the Derivative First, let's find the derivative of Y with respect to x . We have: d x d Y = d x d ( csc x − 2 cos x ) Using the properties of derivatives, we can write: d x d Y = d x d ( csc x ) − 2 d x d ( cos x ) We know that d x d ( csc x ) = − csc x cot x and d x d ( cos x ) = − sin x . Substituting these into the expression, we get: d x d Y = − csc x cot x − 2 ( − sin x ) = − csc x cot x + 2 sin x
Evaluating the Derivative Now, we need to evaluate the derivative at x = 6 π . We have: d x d Y x = 6 π = − csc ( 6 π ) cot ( 6 π ) + 2 sin ( 6 π ) Recall that csc ( 6 π ) = 2 , cot ( 6 π ) = 3 , and sin ( 6 π ) = 2 1 . Substituting these values, we get: d x d Y x = 6 π = − ( 2 ) ( 3 ) + 2 ( 2 1 ) = − 2 3 + 1 = 1 − 2 3
Final Answer Therefore, the rate of change of Y with respect to X at x = 6 π is 1 − 2 3 .
Examples
Understanding rates of change is crucial in many real-world applications. For example, in physics, it helps determine the velocity of an object, which is the rate of change of its position with respect to time. In economics, it can be used to analyze how quickly a company's revenue is changing over time, helping to make informed business decisions. Similarly, in engineering, understanding the rate of change of stress on a material can help prevent structural failures. This problem demonstrates a fundamental concept in calculus that has broad applications across various fields.
