Problem 11 - APEX Calculus

Question

In Exercises \(7-12,\) functions \(z=f(x, y), x=g(t)\) and \(y=h(t)\) are given. (a) Use the Multivariable Chain Rule to compute \(\frac{d z}{d t}\). (b) Evaluate \(\frac{d z}{d t}\) at the indicated \(t\) -value. $$ z=x^{2}+2 y^{2}, \quad x=\sin t, \quad y=3 \sin t ; \quad t=\pi / 4 $$

Step-by-Step Solution

Answer

\(\frac{dz}{dt}\) at \(t=\frac{\pi}{4}\) is 7.

Step 1: Identify the Functions

The given functions are: \[ z = x^2 + 2y^2 \]\[ x = \sin t \]\[ y = 3 \sin t \]We will use these functions to apply the chain rule.

Step 2: Apply the Multivariable Chain Rule

The multivariable chain rule states:\[ \frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt} \]Calculate the required partial derivatives and derivatives.

Step 3: Calculate Partial Derivatives

Compute the partial derivatives of \( z \):\[ \frac{\partial z}{\partial x} = 2x \]\[ \frac{\partial z}{\partial y} = 4y \]

Step 4: Calculate Derivatives of x and y with respect to t

Compute the derivatives:\[ \frac{dx}{dt} = \cos t \]\[ \frac{dy}{dt} = 3 \cos t \]

Step 5: Substitute Values into Chain Rule Equation

Substitute the derivatives into the chain rule equation:\[ \frac{dz}{dt} = (2x)(\cos t) + (4y)(3 \cos t) \]

Step 6: Simplify the Expression

Substitute \( x = \sin t \) and \( y = 3 \sin t \) into the expression:\[ \frac{dz}{dt} = 2(\sin t)(\cos t) + 4(3 \sin t)(3 \cos t) \]\[ = 2 \sin t \cos t + 12 \sin t \cos t \]

Step 7: Factor and Combine Terms

Combine terms:\[ \frac{dz}{dt} = 14 \sin t \cos t \]

Step 8: Evaluate at t = π/4

Substitute \( t = \frac{\pi}{4} \) into the expression:\[ \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2} \] and \[ \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} \]\[ \frac{dz}{dt} = 14 \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{2}}{2}\right) = 14 \left(\frac{2}{4}\right) = 14 \left(\frac{1}{2}\right) = 7 \]

Step 9: Conclusion

Therefore, the value of \( \frac{dz}{dt} \) at \( t = \frac{\pi}{4} \) is 7.

Key Concepts

Chain Rule

The chain rule is an essential tool in calculus, especially when dealing with functions of multiple variables. It allows us to compute the derivative of a composition of functions. In the context of multivariable calculus, the chain rule helps us find the rate at which a function changes with respect to another variable, even when the function is defined in terms of multiple intermediate variables.

In this exercise, we have a function \(z\) that is dependent on two other functions, \(x\) and \(y\), which themselves are dependent on \(t\). The multivariable chain rule helps in determining \(\frac{dz}{dt}\):

• It ensures we account for how both \(x\) and \(y\) change concerning \(t\).
• It calculates the total derivative by combining the partial derivatives with the rates of change of \(x\) and \(y\).

The formula used is: \[ \frac{dz}{dt} = \frac{\partial z}{\partial x} \frac{dx}{dt} + \frac{\partial z}{\partial y} \frac{dy}{dt} \] This approach ensures that all paths of dependence are considered when evaluating the derivative.

Partial Derivatives

Partial derivatives are a way to measure how a multivariable function changes as just one of its variables changes, keeping all other variables constant. They provide insight into the behavior of each individual variable independently of others in the function.

In this exercise:

• The function \(z = x^2 + 2y^2\) requires us to find the rate of change with respect to \(x\) and \(y\) individually.
• The partial derivative of \(z\) with respect to \(x\) is \( \frac{\partial z}{\partial x} = 2x \).
• The partial derivative of \(z\) with respect to \(y\) is \( \frac{\partial z}{\partial y} = 4y \).

These partial derivatives help determine how each variable separately contributes to the overall change in \(z\). They are crucial components of the chain rule when calculating \(\frac{dz}{dt}\).

Trigonometric Functions

Trigonometric functions, such as sine and cosine, regularly appear in calculus problems, especially those involving periodic or oscillatory phenomena. Understanding their derivatives and behavior is crucial.

In this problem:

• The variable \(x = \sin t\), which means it oscillates between -1 and 1 as \(t\) varies.
• Similarly, \(y = 3\sin t\) also oscillates, but with a larger amplitude due to the multiplier 3, ranging from -3 to 3.

The derivatives of sine and cosine are vital:

• The derivative of \(x\) with respect to \(t\) is \( \frac{dx}{dt} = \cos t \).
• The derivative of \(y\) is \( \frac{dy}{dt} = 3\cos t \).

These derivatives play a key role in applying the chain rule, as they describe how quickly \(x\) and \(y\) are changing, given a change in \(t\). Understanding these trigonometric derivatives helps greatly in problems involving periodic functions.

Derivative Evaluation

Derivative evaluation is where all the hard work of calculating derivatives and partial derivatives pays off. This involves substituting specific values into the derivatives we have computed to find a numerical result. In this exercise, we are asked to evaluate \(\frac{dz}{dt}\) at a specific point \(t = \pi/4\).

Here is how it is done:

• First, calculate \(\sin(\pi/4)\) and \(\cos(\pi/4)\), both equal to \(\frac{\sqrt{2}}{2}\).
• Substitute these values back into the expression for \(\frac{dz}{dt}\). The simplified expression \(14 \sin t \cos t\) becomes \(14 \left(\frac{\sqrt{2}}{2}\right)^2\).
• Finally, simplify this result to obtain \(\frac{dz}{dt} = 7\) at \(t = \pi/4\).

This step confirms the precise rate of change of \(z\) with respect to \(t\) at the specific angle, illustrating the power of derivative evaluation in providing concrete results.