Problem 8 - Calculus Early Transcendentals

Question

Use Theorem 7 to find the following derivatives. When feasible, express your answer in terms of the independent variable. $$d z / d t, \text { where } z=x y^{2}, x=t^{2}, \text { and } y=t$$

Step-by-Step Solution

Answer

Question: Find the derivative of \(z(x(t), y(t))\) with respect to the independent variable \(t\), given that \(z(x, y) = x y^2\), \(x(t) = t^2\), and \(y(t) = t\). Answer: The derivative of \(z(x(t), y(t))\) with respect to the independent variable \(t\) is \(\frac{dz}{dt} = 2t^3 + 2t^4\).

Step 1: Find the partial derivatives of z

To find the partial derivatives of z with respect to x and y, we will differentiate z with respect to x while keeping y constant and differentiate z with respect to y while keeping x constant. $$\frac{\partial z}{\partial x} = \frac{\partial (xy^2)}{\partial x} = y^2$$ $$\frac{\partial z}{\partial y} = \frac{\partial (xy^2)}{\partial y} = 2xy$$

Step 2: Find the derivatives of x(t) and y(t)

Now, we will compute the derivatives of \(x(t)\) and \(y(t)\) with respect to \(t\). $$\frac{dx}{dt} = \frac{d (t^2)}{dt} = 2t$$ $$\frac{dy}{dt} =\frac{d (t)}{dt} = 1$$

Step 3: Apply the Chain Rule to find dz/dt

By applying the Chain Rule, we can find the derivative of z with respect to t as follows: $$\frac{dz}{dt} = \frac{\partial z}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial z}{\partial y} \cdot \frac{dy}{dt}$$ Now, substitute the partial derivatives and the derivatives of x(t) and y(t) that we found in steps 1 and 2: $$\frac{dz}{dt} = (y^2) \cdot (2t) + (2xy) \cdot (1)$$

Step 4: Express the answer in terms of the independent variable t

Finally, we need to express the result in terms of the independent variable t. Since \(x(t) = t^2\) and \(y(t) = t\), we can replace x and y with their respective expressions in terms of t: $$\frac{dz}{dt} = (t^2) \cdot (2t) + (2(t^2)t) \cdot (1)$$ Simplify the expression: $$\frac{dz}{dt} = 2t^3 + 2t^4$$ Now, we have found the derivative of z with respect to the independent variable t: $$\frac{dz}{dt} = 2t^3 + 2t^4$$

Key Concepts

Partial Derivatives

A partial derivative is used when a function has more than one variable and we want to determine how the function changes with respect to just one of these variables, while keeping the others constant. In the context of the given exercise, the function is defined as \(z = xy^2\). Here, \(z\) is dependent on both \(x\) and \(y\), making this a perfect scenario for using partial derivatives.
To find the partial derivative of \(z\) with respect to \(x\), denoted as \(\frac{\partial z}{\partial x}\), we treat \(y\) as a constant and differentiate \(z\) with respect to \(x\). This yields \(y^2\), because the only variable affecting \(z\) in this term is \(x\). Similarly, when taking the partial derivative with respect to \(y\), \(x\) is held constant. Thus, \(\frac{\partial z}{\partial y} = 2xy\), because this time \(y\) is directly multiplying with \(2y\), producing the term \(2xy\).
- **Why Partial Derivatives?** - They allow us to understand the relationship between specific variables. - They are crucial when dealing with multivariate functions, like in this problem.

Differentiation

Differentiation is a fundamental concept in calculus that deals with finding how a function changes as its input changes. In simpler terms, it involves finding the rate of change or the slope of a function. For our exercise, differentiation is applied twice:

• First, to find how \(x(t)\) and \(y(t)\) change with respect to time \(t\).
• Second, through the concept of the chain rule in the final step.

When we compute \(\frac{dx}{dt}\) for \(x = t^2\), we find it equals \(2t\), because \(x\) changes at a rate proportional to \(t\). For \(y = t\), \(\frac{dy}{dt}\) is 1, as \(y\) increases linearly with \(t\).
Understanding differentiation as identifying the slope or gradient helps in predicting future values of the function. Differentiation forms the core for predictive modeling in many fields, such as physics and economics.

Independent Variable

The independent variable in a function is the variable that can be freely changed, allowing us to see how it affects the dependent variables. In this exercise, \(t\) is defined as the independent variable. Both \(x\) and \(y\) are expressed in terms of \(t\) (i.e., \(x = t^2\) and \(y = t\)).
This framework is essential because it sets up how the functions \(x\) and \(y\) evolve over time \(t\). By expressing everything in terms of \(t\), the problem becomes more manageable, as we've isolated one variable to focus on.
In essence:

• The independent variable is like a controlling knob that influences how the other variables in the equation behave.
• Expressing derivatives in terms of this variable simplifies complex problems into more digestible steps, conducive to clearer solutions.

Understanding the role of an independent variable is crucial for both mathematical problems and real-world applications.