Problem 14 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Answer: The derivative of the function Q with respect to time t is $\frac{dQ}{dt} = 1$.

Step 1: Find the partial derivatives of Q

To begin, we will differentiate Q with respect to each of the three variables x, y, and z to find the partial derivatives. We have $$Q=\sqrt{x^{2}+y^{2}+z^{2}}$$ Taking the partial derivatives with respect to x, y, and z, we get: $$\begin{aligned} \frac{\partial Q}{\partial x} & = \frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}} \\ \frac{\partial Q}{\partial y} & = \frac{y}{\sqrt{x^{2}+y^{2}+z^{2}}} \\ \frac{\partial Q}{\partial z} & = \frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}} \end{aligned}$$

Step 2: Find the time derivatives of x, y, and z

Now we will find the derivatives of x, y, and z with respect to time t. We are given: $$\begin{aligned} x & = \sin t \\ y & = \cos t \\ z & = \cos t \end{aligned}$$ Differentiating each with respect to t, we get: $$\begin{aligned} \frac{dx}{dt} & = \cos t \\ \frac{dy}{dt} & = -\sin t \\ \frac{dz}{dt} & = -\sin t \end{aligned}$$

Step 3: Apply the Chain Rule

Now we will apply the Chain Rule (Theorem 7) to find the derivative of Q with respect to t. We have: $$\frac{dQ}{dt} = \frac{\partial Q}{\partial x}\frac{dx}{dt} + \frac{\partial Q}{\partial y}\frac{dy}{dt} + \frac{\partial Q}{\partial z}\frac{dz}{dt}$$ Substitute the values we found in Steps 1 and 2: $$\frac{dQ}{dt} = \frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}}\cos t + \frac{y}{\sqrt{x^{2}+y^{2}+z^{2}}}(-\sin t) + \frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}}(-\sin t)$$ Now substitute the original expressions for x, y, and z in terms of t: $$\frac{dQ}{dt} = \frac{\sin t}{\sqrt{(\sin t)^{2}+(\cos t)^{2}+(\cos t)^{2}}}\cos t - \frac{\cos t}{\sqrt{(\sin t)^{2}+(\cos t)^{2}+(\cos t)^{2}}}\sin t - \frac{\cos(t)}{\sqrt{(\sin t)^{2}+(\cos t)^{2}+(\cos t)^{2}}} \sin t$$

Step 4: Simplify the expression

Simplify the expression by factoring out the common term in the denominator and cancelling terms: $$\frac{dQ}{dt} = \frac{\sqrt{(\sin t)^{2}+(\cos t)^{2}+(\cos t)^{2}}}{\sqrt{(\sin t)^{2}+(\cos t)^{2}+(\cos t)^{2}}}$$ Finally, we have: $$\frac{dQ}{dt} = 1$$

Key Concepts

Partial Derivatives

In multivariable calculus, a partial derivative represents how a function changes as only one of the variables is changed slightly, while all other variables are held constant. This is a crucial concept when dealing with functions of several variables, like in the original exercise involving function \( Q = \sqrt{x^2 + y^2 + z^2} \).
This function depends on three variables \( x \), \( y \), and \( z \) and requires finding individual derivatives—called partial derivatives—for each variable.
When calculating the partial derivative of \( Q \) with respect to \( x \), for example, you treat \( y \) and \( z \) as constants and differentiate only with respect to \( x \). The formula for this becomes \( \frac{\partial Q}{\partial x} = \frac{x}{\sqrt{x^2 + y^2 + z^2}} \).
Similarly, you apply the same process to find \( \frac{\partial Q}{\partial y} \) and \( \frac{\partial Q}{\partial z} \).
This process of looking at how a function's output changes with the change in just one input while keeping others fixed is what makes partial derivatives incredibly useful in multivariable calculus.

Multivariable Calculus

Multivariable calculus extends calculus concepts, such as differentiation and integration, to functions of several variables. In the original problem, the function \( Q = \sqrt{x^2 + y^2 + z^2} \) is a function in three-dimensional space and such problems are typical in multivariable calculus.
When you work with these types of functions, understanding how they change with respect to each variable helps unravel complex systems in fields like physics and engineering.
Here, we deal with changing functions along multiple axes simultaneously and the relationships among these changes.

• In two or more dimensions, the concept of a derivative is extended to partial derivatives for each variable.
• This allows us to describe surfaces or, more generally, hypersurfaces in higher-dimensional spaces.

The exercise also illustrates using multiple partial derivatives and the Chain Rule to find how the function \( Q \) changes with time, incorporating the relationships between variables as a function of another parameter, \( t \). This is precisely the power of multivariable calculus—unlocking the potential to model and problem-solve in real-world multi-faceted scenarios.

Parametric Derivatives

Parametric derivatives arise when you have variables that are expressed parametrically in terms of another variable. In other words, functions like \( x = \sin t \), \( y = \cos t \), and \( z = \cos t \) are written as functions of another variable \( t \).
In the exercise, each of the variables \( x \), \( y \), and \( z \) is expressed in terms of \( t \), which is common in physical scenarios like motion along paths where time is the parameter.
To find the derivative of \( Q \) with respect to \( t \), each parameter's change concerning \( t \) needs to be accounted for, which is solved using the Chain Rule.

• We first compute each derivative, such as \( \frac{dx}{dt} = \cos t \).
• Then, these derivatives are combined with the partial derivatives using the relation \( \frac{dQ}{dt} = \frac{\partial Q}{\partial x}\frac{dx}{dt} + \frac{\partial Q}{\partial y}\frac{dy}{dt} + \frac{\partial Q}{\partial z}\frac{dz}{dt} \).

This approach uncovers how the derivative of a parametric function encompasses both the rates of change of the individual variables and the interconnection through the parameter \( t \). Using parametric derivatives, complex systems reliant on time as a driving factor can be meticulously understood and navigated.