Question

Find \(\partial w / \partial t\) by using the Chain Rule. Express your final answer in terms of \(s\) and \(t\) $$ w=\sqrt{x^{2}+y^{2}+z^{2}}, x=\cos s t, y=\sin s t, z=s^{2} t $$

Short answer

\( \frac{\partial w}{\partial t} = \frac{s^4 t}{\sqrt{1 + s^4 t^2}} \).

Step-by-step solution

Identify the Dependencies

The problem involves finding the partial derivative \( \frac{\partial w}{\partial t} \) where \( w = \sqrt{x^2 + y^2 + z^2} \). The variables \( x, y, \) and \( z \) are functions of \( s \) and \( t \); specifically, \( x = \cos(st) \), \( y = \sin(st) \), and \( z = s^2 t \).

Apply the Chain Rule

The Chain Rule for partial derivatives states that: \[\frac{\partial w}{\partial t} = \frac{\partial w}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial w}{\partial y} \cdot \frac{\partial y}{\partial t} + \frac{\partial w}{\partial z} \cdot \frac{\partial z}{\partial t}\] We will need to determine each partial derivative separately.

Compute Partial Derivatives for \(w\)

First, find the partial derivatives of \( w \) with respect to \( x, y, \) and \( z \):\[\frac{\partial w}{\partial x} = \frac{x}{\sqrt{x^2 + y^2 + z^2}}, \quad \frac{\partial w}{\partial y} = \frac{y}{\sqrt{x^2 + y^2 + z^2}}, \quad \frac{\partial w}{\partial z} = \frac{z}{\sqrt{x^2 + y^2 + z^2}}\] These derivatives follow from the chain rule applied to \( w = \sqrt{x^2 + y^2 + z^2} \).

Compute Partial Derivatives for \(x, y,\) and \(z\)

Now, compute the partial derivatives of \( x, y, \) and \( z \) with respect to \( t \):\[\frac{\partial x}{\partial t} = -s \sin(st), \quad \frac{\partial y}{\partial t} = s \cos(st), \quad \frac{\partial z}{\partial t} = s^2\]These results are derived from the standard differentiation rules applied to the given functions of \( t \).

Substitute and Simplify

Substitute the derivatives into the chain rule expression:\[\frac{\partial w}{\partial t} = \frac{x}{\sqrt{x^2 + y^2 + z^2}} (-s \sin(st)) + \frac{y}{\sqrt{x^2 + y^2 + z^2}} (s \cos(st)) + \frac{z}{\sqrt{x^2 + y^2 + z^2}} (s^2)\]This simplifies to:\[\frac{\partial w}{\partial t} = \frac{-x s \sin(st) + y s \cos(st) + zs^2}{\sqrt{x^2 + y^2 + z^2}}\] Due to the identities in trigonometric differentiation, recognize that the components in \( x \) and \( y \) terms simplify further.

Final Simplification

Insert \( x = \cos(st), y = \sin(st), z = s^2t \) into the equation:\[\frac{\partial w}{\partial t} = \frac{-s \cos(st) \sin(st) + s \sin(st) \cos(st) + s^2(s^2 t)}{\sqrt{\cos^2(st) + \sin^2(st) + (s^2 t)^2}}\]Utilize the identity \( \cos^2(st) + \sin^2(st) = 1 \) for further simplification:\[\frac{\partial w}{\partial t} = \frac{s^2(s^2 t)}{\sqrt{1 + (s^2 t)^2}} = \frac{s^4 t}{\sqrt{1 + s^4 t^2}}\] This is the expression fully simplified in terms of \( s \) and \( t \).

Key concepts

Partial Derivatives

Partial derivatives allow us to understand how functions change as one of their variables changes, while keeping other variables constant. In multivariable calculus, we often deal with functions that depend on several variables. Here, the function is represented as \( w = \sqrt{x^2 + y^2 + z^2} \), where \( x, y, \) and \( z \) themselves depend on variables \( s \) and \( t \).
This makes \( w \) a multivariable function since it eventually depends on both \( s \) and \( t \) through its dependency on \( x, y, \) and \( z \).
Using the chain rule in the context of partial derivatives allows us to calculate how \( w \) changes with a small change in \( t \), denoted as \( \frac{\partial w}{\partial t} \).
This involves finding partial derivatives of \( w \) with respect to each of the intermediate variables \( x, y, \) and \( z \) and then combining these with the derivatives of \( x, y, \) and \( z \) with respect to \( t \).
By doing this, we can capture the full effect of \( t \) on \( w \), providing a deeper understanding of the system's dynamic behavior.

Multivariable Functions

Multivariable functions, such as \( w = \sqrt{x^2 + y^2 + z^2} \), are functions with more than one input variable. In this context:

• The function \( w \) indirectly depends on the variables \( s \) and \( t \) through \( x = \cos(st), y = \sin(st), \) and \( z = s^2 t \).
• By expressing these dependencies, we understand how interconnected variables can describe complex relationships.

When dealing with such functions, it is important to carefully establish how each variable relates to others, as this impacts the calculation of derivatives.
The complexities of multivariable functions enhance our understanding of real-world phenomena, where one outcome may result from many different influences.
In mathematical modeling, defining clear variable dependencies and building a structured approach to differentiate with respect to parameters like \( t \) is crucial.
This ensures we capture vital interactions between variables accurately.

Trigonometric Differentiation

Trigonometric differentiation involves computing derivatives for functions incorporating trigonometric operations, like \( \cos \) and \( \sin \).
In our case:

• For \( x = \cos(st) \), the derivative with respect to \( t \) is calculated using the product rule, leading to \(-s \sin(st)\). This is because \( \frac{d(\cos u)}{du} = -\sin u \) and \( u = st \).
• For \( y = \sin(st) \), the derivative becomes \( s \cos(st) \), since \( \frac{d(\sin u)}{du} = \cos u \).

Trigonometric identities and differentiation rules are foundational in applying calculus to trigonometric components within larger functions.
Understanding these derivatives lets us observe how changes in angle or phase, altered by variables such as \( s \) and \( t \), influence our function's behavior.
By appropriately handling trigonometric differentiation, we translate these angular relationships into derivatives that can portray variations in intricate systems effectively.

Calculus Problem Solving

Solving calculus problems, like finding \( \frac{\partial w}{\partial t} \), involves systematically applying mathematical principles and rules. Here's the process:

• First, understand the dependencies within the function, recognizing how components like \( x, y, \) and \( z \) depend on \( s \) and \( t \).
• Next, apply the chain rule to determine the overall effect of changing \( t \) on \( w \), incorporating individual partial derivatives.
• Compute the necessary derivatives of all intermediate variables based on known rules, such as trigonometric differentiation.

Finally, substitute and simplify your expressions to form a clear, precise result.
This structured method showcases the power of calculus in breaking down complex relationships into understandable parts.
Through practice, developing problem-solving skills in calculus ensures easier tackling of diverse problems, building a solid foundation for understanding dynamic systems based on logical deduction and analysis.