Problem 215 - Calculus 3

Question

For the following exercises, use the information provided to solve the problem. Let \(w(x, y, z)=x y \cos z, \quad\) where \(\quad x=t, y=t^{2}\) and \(z=\arcsin t .\) Find \(\frac{d w}{d t}\)

Step-by-Step Solution

Answer

\(\frac{dw}{dt} = \frac{3t^2 - 4t^4}{\sqrt{1-t^2}}\).

Step 1: Identify the Problem Type

The problem involves finding the derivative of a function of multiple variables \(w(x, y, z)\) where each variable \(x\), \(y\), and \(z\) is in terms of another variable \(t\). This is a classic example of using the chain rule in partial differentiation.

Step 2: Understand Function Composition

We have \(w(x, y, z) = xy \cos z\) where it is provided that \(x = t\), \(y = t^2\), and \(z = \arcsin t\). Hence, \(w\) is a composite function depending on \(t\) through \(x\), \(y\), and \(z\).

Step 3: Apply the Chain Rule

The derivative \(\frac{dw}{dt}\) can be found using the chain rule for multi-variable functions:\[\frac{dw}{dt} = \frac{\partial w}{\partial x}\frac{dx}{dt} + \frac{\partial w}{\partial y}\frac{dy}{dt} + \frac{\partial w}{\partial z}\frac{dz}{dt}.\]

Step 4: Calculate Partial Derivatives

Find each of the partial derivatives:- \(\frac{\partial w}{\partial x} = y \cos z\),- \(\frac{\partial w}{\partial y} = x \cos z\),- \(\frac{\partial w}{\partial z} = -xy \sin z\).

Step 5: Calculate Derivatives of Intermediate Variables

Derive each variable with respect to \(t\):- \(\frac{dx}{dt} = \frac{d(t)}{dt} = 1\),- \(\frac{dy}{dt} = \frac{d(t^2)}{dt} = 2t\),- \(\frac{dz}{dt} = \frac{d(\arcsin t)}{dt} = \frac{1}{\sqrt{1-t^2}}\).

Step 6: Substitute and Simplify

Substitute the results from steps 4 and 5 into the chain rule expression:\[\frac{dw}{dt} = (y \cos z)(1) + (x \cos z)(2t) - (xy \sin z)\left(\frac{1}{\sqrt{1-t^2}}\right).\]Substitute \(x = t\), \(y = t^2\), and \(z = \arcsin t\), giving:\[\frac{dw}{dt} = (t^2 \cos(\arcsin t)) + (t \cos(\arcsin t))(2t) - (t \cdot t^2 \sin(\arcsin t))\left(\frac{1}{\sqrt{1-t^2}}\right).\]

Step 7: Simplify Trigonometric Expressions

Recall trigonometric identities:- \(\cos(\arcsin t) = \sqrt{1-t^2}\),- \(\sin(\arcsin t) = t\).Using these identities:\[\frac{dw}{dt} = (t^2 \sqrt{1-t^2}) + (2t^2 \sqrt{1-t^2}) - \frac{t^3 t}{\sqrt{1-t^2}}.\]

Step 8: Combine Terms

Combine terms into a common expression:\[\frac{dw}{dt} = (t^2 + 2t^2)\sqrt{1-t^2} - \frac{t^4}{\sqrt{1-t^2}}.\]Combine further by simplifying:\[\frac{dw}{dt} = 3t^2 \sqrt{1-t^2} - \frac{t^4}{\sqrt{1-t^2}} = \frac{3t^2(1-t^2) - t^4}{\sqrt{1-t^2}}.\]After distributing, the expression is simplified to:\[\frac{dw}{dt} = \frac{3t^2 - 3t^4 - t^4}{\sqrt{1-t^2}} = \frac{3t^2 - 4t^4}{\sqrt{1-t^2}}.\]

Key Concepts

Partial Derivatives

When dealing with functions of multiple variables, partial derivatives help us understand how the function changes with respect to one variable while keeping others constant. In our exercise, the function \(w(x, y, z)\) depends on three variables: \(x\), \(y\), and \(z\). We found partial derivatives \(\frac{\partial w}{\partial x}\), \(\frac{\partial w}{\partial y}\), and \(\frac{\partial w}{\partial z}\). These measure:

• How \(w\) changes when \(x\) changes but \(y\) and \(z\) remain fixed.
• How \(w\) changes with changes in \(y\) only.
• How \(w\) changes exclusively as \(z\) changes.

This is crucial for applying the chain rule later. Calculating these derivatives gives insight into the individual impact each variable has on \(w\).

Composite Functions

Composite functions involve functions within functions. In this exercise, \(w = x y \cos z\) not only relies on \(x\), \(y\), and \(z\), but these are further dependent on \(t\). This layering requires understanding how modification in \(t\) affects \(x\), \(y\), and \(z\), and subsequently \(w\) itself.Understanding them begins with identifying the inner functions, \(x = t\), \(y = t^2\), and \(z = \arcsin t\). Each is directly influenced by the variable \(t\), combining to influence \(w\) indirectly through these relationships. Calculating derivatives of each intermediate function with respect to \(t\) helps in using the chain rule to find \(\frac{d w}{d t}\).

Trigonometric Identities

Trigonometric identities are vital in simplifying expressions involving trigonometric functions. In the context of this problem, simplifying \(\cos(\arcsin t)\) and \(\sin(\arcsin t)\) is crucial.To simplify:

• \(\cos(\arcsin t) = \sqrt{1 - t^2}\).
• \(\sin(\arcsin t) = t\).

These simplifications turn complex-looking expressions into more manageable forms. Recognizing and applying these identities reduces the burden of trigonometric expressions during calculations, enabling easier handling of complex derivatives.

Multivariable Calculus

Multivariable calculus extends calculus concepts to functions involving multiple inputs. Finding \(\frac{d w}{d t}\) in our exercise showcases how to handle functions with variables relying on another variable \(t\).In multivariable calculus:

• Each variable can vary independently contributing differently to the overall derivative.
• The chain rule connects these independent variables back to a single derivative with \(t\).

This exercise required combining all previously calculated partial derivatives and intermediate derivative values. It illustrates the power of multivariable calculus, allowing us to navigate the multi-layered influence of individual variables to determine how a primary function changes with respect to a single underpinning variable. Understanding these principles is key in fields where outcomes depend on several related factors.