Problem 3 - Calculus Single Variable

Question

The variable \(y\) is given as a function of \(x\), which depends on \(t\). The values \(x_{0}\) and \(v_{0}\) of, respectively, \(x\) and \(d x / d t\) are given at a value \(t_{0}\) of \(t\). Use this data to find \(d y / d t\) at \(t_{0}\). $$ y=\cos (x), \quad x_{0}=\pi / 6, \quad v_{0}=-2 $$

Step-by-Step Solution

Answer

\( \frac{dy}{dt} \) at \( t_0 \) is 1.

Step 1: Understand the Given Problem

We are given the function \( y = \cos(x) \) where \( y \) depends on \( x \), and \( x \) depends on \( t \). We need to find \( \frac{dy}{dt} \). Also, we know \( x_0 = \frac{\pi}{6} \) and \( \frac{dx}{dt} \) or \( v_0 = -2 \) at \( t = t_0 \).

Step 2: Differentiate y with respect to t using the Chain Rule

Since \( y = \cos(x) \) and \( x \) is a function of \( t \), we apply the chain rule to differentiate \( y \) with respect to \( t \). This gives us:\[\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt} = -\sin(x) \cdot \frac{dx}{dt}\]

Step 3: Substitute Known Values into the Derivative

Substitute \( x = x_0 = \frac{\pi}{6} \) and \( \frac{dx}{dt} = v_0 = -2 \) into the expression for \( \frac{dy}{dt} \):\[\frac{dy}{dt} = -\sin\left(\frac{\pi}{6}\right) \cdot (-2)\]

Step 4: Calculate \( \sin(\frac{\pi}{6}) \)

The value of \( \sin(\frac{\pi}{6}) \) is known to be \( \frac{1}{2} \).

Step 5: Compute \( \frac{dy}{dt} \)

Substitute \( \sin(\frac{\pi}{6}) = \frac{1}{2} \) back into the equation for \( \frac{dy}{dt} \):\[\frac{dy}{dt} = -\frac{1}{2} \cdot (-2) = 1\]

Key Concepts

Implicit Differentiation

When working with functions where one variable depends on another, and those variables are connected through an intermediate variable, we use implicit differentiation to find the derivatives. In the given exercise, we've a function:

• \( y = \cos(x) \)
• x is a function of t

This is implicit because y is not given directly as a function of t, but instead through x.
To differentiate \( y = \cos(x) \) with respect to t, we cannot do it directly due to the indirect dependency through x. This is where the Chain Rule comes in to connect the dots. Implicit differentiation helps by considering both the roles of x and t in affecting y's behavior.
By applying implicit differentiation, we recognize that we need to use the derivative of y with respect to x and link it to the derivative of x with respect to t, hence linking y indirectly back to t.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are fundamental in differentiating and solving calculus problems involving rates of change. In this context, we are working with the function \( y = \cos(x) \).
The derivative of \( \cos(x) \) with respect to x is \( -\sin(x) \). This result is critical when differentiating y implicitly with respect to t, using the Chain Rule as mentioned before.
Understanding the properties of trigonometric functions is essential. Specifically, know the standard values such as \( \sin(\frac{\pi}{6}) = \frac{1}{2} \) and \( \cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2} \). Such known values simplify calculus steps tremendously when solving problems.
In our example, identifying \( \sin(\frac{\pi}{6}) \) as \( \frac{1}{2} \) helps complete the differentiation process accurately.

Calculus Problem Solving

Solving calculus problems like the given exercise involves both identifying relationships and applying the appropriate rules. In this exercise:

• The Chain Rule is used to differentiate the function \( y = \cos(x) \) with respect to t.
• An understanding of derivative properties of trigonometric functions is needed.

To solve for \( \frac{dy}{dt} \), start by recognizing the need for the Chain Rule due to the intermediate dependency of x on t. Once the relationships are established, substitute known values to simplify the expression.
Use the formula given:\[\frac{dy}{dt} = -\sin(x) \cdot \frac{dx}{dt}\]Substituting \( x = \frac{\pi}{6} \) and \( \frac{dx}{dt} = -2 \), the equation becomes:\[\frac{dy}{dt} = -\frac{1}{2} \cdot (-2)\]Calculating, you find that \( \frac{dy}{dt} = 1 \).
This is a step-by-step application of implicit differentiation and trigonometric identities. Calculus problem-solving typically integrates multiple mathematical strategies, like substitution and derivative calculations, to find solutions.