Problem 215 - Calculus Volume 3

Question

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

Step-by-Step Solution

Answer

\(\frac{d w}{d t} = \frac{2t^2 - 3t^4}{\sqrt{1-t^2}}\)

Step 1: Understand the problem

We have a function \(w(x, y, z) = x y \cos{z}\), where \(x = t\), \(y = t^2\), and \(z = \arcsin{t}\). We need to find \(\frac{d w}{d t}\). This involves applying the chain rule for differentiating a function of multiple variables with respect to \(t\).

Step 2: Express w in terms of t

Write \(w\) purely in terms of \(t\), using given expressions: \(w(t) = t \cdot t^2 \cdot \cos(\arcsin{t}) = t^3 \cdot \cos(\arcsin{t})\).

Step 3: Differentiate w(t) with respect to t

Apply the product rule \(\frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t)\) to \(w(t) = t^3 \cos(\arcsin{t})\). Define \(u(t) = t^3\) and \(v(t) = \cos(\arcsin{t})\).

Step 4: Differentiate u(t)

Calculate \(u'(t) = \frac{d}{dt} [t^3] = 3t^2\).

Step 5: Differentiate v(t)

Express \(v(t) = \cos(\arcsin{t})\). Using the identity \(\sin^2 z + \cos^2 z = 1\), \(\cos^2 z = 1 - \sin^2 z\), where \(\sin z = t\).

Step 6: Simplify cosine using identity

Since \(z = \arcsin t\), we have \(\cos z = \sqrt{1 - t^2}\). Therefore, \(v(t) = \sqrt{1 - t^2}\).

Step 7: Find v'(t)

Differentiate \(v(t) = \sqrt{1 - t^2}\) using chain rule: \(v'(t) = -\frac{t}{\sqrt{1-t^2}}\).

Step 8: Apply product rule

The derivative of \(w(t)\) is given by: \[\frac{d w}{d t} = u'(t) v(t) + u(t) v'(t) = 3t^2 \sqrt{1 - t^2} + t^3 \left(-\frac{t}{\sqrt{1-t^2}}\right)\].

Step 9: Simplify

Combine and simplify: \[\frac{d w}{d t} = 3t^2 \sqrt{1 - t^2} - \frac{t^4}{\sqrt{1-t^2}} = \frac{3t^2(1-t^2) - t^4}{\sqrt{1-t^2}}\].

Key Concepts

Partial Derivatives

When we talk about partial derivatives, we're focusing on functions with more than one variable. In this case, the function \(w(x, y, z) = xy\cos{z}\) involves three variables: \(x, y,\) and \(z\). Partial derivatives help us understand how the function changes with respect to one variable, holding the others constant.

• For example, \(\frac{\partial w}{\partial x}\) analyzes how \(w\) changes just with changes in \(x\), keeping \(y\) and \(z\) stable.
• This is a building block for the chain rule, which allows us to calculate derivatives in more complex situations where all variables may be changing.

In problems like ours, direct partial derivatives might seem out of sight because transformations have already put everything in terms of \(t\). However, understanding these derivatives is crucial for applying the chain rule correctly.

Product Rule

The product rule is a key tool in calculus for differentiating products of two functions. Here, it's used to differentiate \(w(t) = t^3 \cos(\arcsin{t})\). The rule states: \[\frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t)\]

• With \(u(t) = t^3\) and \(v(t) = \cos(\arcsin{t})\), the rule helps split the problem into two simpler derivatives.
• First, differentiate each part separately (finding \(u'(t)\) and \(v'(t)\)).
• Then, combine these using the product rule formula to get the derivative of the entire product.

This approach is essential when a function is a product, making complex differentiations manageable.

Trigonometric Functions

Trigonometric functions play a significant role in our problem, especially \(\cos\) and the inverse \(\arcsin\). The relationship between \(\cos\) and \(\sin\) is governed by the identity \(\sin^2 z + \cos^2 z = 1\).

• If \(z = \arcsin t\), then \(\sin z = t\) and \(\cos z = \sqrt{1 - t^2}\).
• This transformation is crucial because it simplifies the differentiation of \(\cos(\arcsin t)\).
• Understanding these identities allows you to manipulate trig functions and find derivatives or simplify expressions.

Trigonometric identities and functions often appear in calculus problems. Recognizing these helps to simplify many complex derivations, especially those involving inverse trigonometric forms.

Calculus Problem Solving

Solving a calculus problem often involves a structured approach. Here, we've tackled finding \(\frac{d w}{d t}\) for \(w(t) = t^3 \cos(\arcsin t)\). Key steps include:

• Break down the problem: Clearly understand each part of the function and rewrite it if necessary. For instance, \(w\) was initially in terms of \(x, y, z\) and needed expression in \(t\).
• Apply appropriate rules: Utilize derivative rules like the product rule and chain rule to address each component of the function.
• Follow through simplification: After finding individual derivatives, combine them and simplify where possible. Here, combining the result using algebra simplifies the final derivative.

By following a structured approach, calculus problems become less daunting and more methodical, facilitating a clear path to the solution.