Problem 247 - Calculus 3

Question

If \(w=\sin (x y z), x=1-3 t, y=e^{1-t}\) and \(z=4 t, \quad\) find \(\frac{\partial w}{\partial t}\)

Step-by-Step Solution

Answer

The partial derivative \(\frac{\partial w}{\partial t}\) is \(\cos(4t(1-3t)e^{1-t}) \cdot 4e^{1-t}(1-6t+3t^2)\).

Step 1: Understand the function for w

We have the function \(w = \sin(xyz)\), where \(x = 1 - 3t\), \(y = e^{1-t}\), and \(z = 4t\). First, substitute these expressions into the function \(w\).

Step 2: Substitute expressions for x, y, and z into w

Substitute the given expressions for \(x\), \(y\), and \(z\) into \(w\):\[w = \sin\left((1 - 3t) \cdot e^{1-t} \cdot 4t\right)\] Simplifying the product inside the sine function gives us:\[w = \sin(4t(1-3t)e^{1-t})\]

Step 3: Differentiate w with respect to t using the chain rule

To find \(\frac{\partial w}{\partial t}\), apply the chain rule. Differentiate the sine function using \(\frac{d}{dt}[\sin(u)] = \cos(u) \cdot u'\), where \(u = 4t(1-3t)e^{1-t}\).

Step 4: Find the derivative of u with respect to t

We have \(u = 4t(1-3t)e^{1-t}\). Now, use the product rule and chain rule to find \(\frac{du}{dt}\).Find the derivative for each part: 1. \(\frac{d}{dt}[4t] = 4\).2. \(\frac{d}{dt}[1-3t] = -3\).3. \(\frac{d}{dt}[e^{1-t}] = -e^{1-t}\) (chain rule).

Step 5: Apply the product rule to differentiate u

Apply the product rule:\(\frac{du}{dt} = \left(\frac{d}{dt}[4t](1-3t)e^{1-t} + 4t\frac{d}{dt}[1-3t]e^{1-t} + 4t(1-3t)\frac{d}{dt}[e^{1-t}]\right)\).Substitute derivatives from step 4:\[4(1-3t)e^{1-t} - 12te^{1-t} - 4t(1-3t)e^{1-t}\]Simplify to get:\[4e^{1-t}(1-3t - 3t + 3t^2)\].

Step 6: Final computation of partial derivative of w with respect to t

Combine the results from previous steps to get:\[\frac{\partial w}{\partial t} = \cos(4t(1-3t)e^{1-t}) \cdot 4e^{1-t}(1-6t+3t^2)\].

Key Concepts

Chain Rule

The chain rule is a fundamental technique used in calculus for differentiating composite functions. When we have a function nested inside another function, the chain rule helps us find the derivative. In mathematical terms, if you have a function \( w = f(g(t)) \), the chain rule states that the derivative \( \frac{dw}{dt} \) is \( \frac{dw}{dg} \cdot \frac{dg}{dt} \). This means you differentiate the outer function \( f \) with respect to \( g \), and multiply it by the derivative of the inner function \( g \) with respect to \( t \).

In our exercise, the chain rule enabled us to differentiate \( w = \sin(u) \) with respect to \( t \). Here, \( u \) is a complex function of \( t \) as well, given by \( u = 4t(1-3t)e^{1-t} \). The chain rule guided us in first finding \( \frac{d \sin(u)}{du} = \cos(u) \) and then multiplying it by the derivative of \( u \) with respect to \( t \).

Understanding the chain rule empowers you to handle similar challenges where functions sit atop layers of other functions. It's like peeling an onion, solving it layer by layer!

Product Rule

The product rule is essential when you're dealing with products of two or more functions. It's particularly useful when you need to differentiate a function composed of products of simpler functions.
In its basic form, the product rule states that for two functions \( u(t) \) and \( v(t) \), the derivative of their product with respect to \( t \) is given by:

• \( \frac{d}{dt}[u(t)v(t)] = u'(t)v(t) + u(t)v'(t) \)

The rule tells us to differentiate the first function and multiply by the second, then add the product of the first function with the derivative of the second function.

In our specific case, we applied the product rule to \( u = 4t(1-3t)e^{1-t} \), which is a product of three functions \( 4t \), \( (1-3t) \), and \( e^{1-t} \). We broke it down systematically, distinguishing how each part contributes to the entire derivative. Breaking it into smaller parts, calculating each derivative separately, and assembling them gives clarity to our solution approach.

Understanding the product rule not only gives you the tools to handle complex derivatives, but it also enhances your problem-solving skills as you learn to navigate functions that interact multiplicatively.

Trigonometric Functions

Trigonometric functions, like sine and cosine, show up frequently in calculus problems, including our exercise where \( w = \sin(xyz) \). When differentiating sine and cosine, their unique properties must be remembered:

• The derivative of \( \sin(u) \) is \( \cos(u) \).
• The derivative of \( \cos(u) \) is \(-\sin(u) \).

These derivatives reflect the fact that sine and cosine are continuous wave functions, crucial for physics-dependent calculations and engineering applications.

In the context of partial derivatives, we focused on the derivative of \( \sin(u) \) while holding other variables constant or treating them singularly due to their dependence on \( t \). By doing so, we ensure that our solution considers how changes in \( t \) ripple through to affect \( w \).
Trigonometric functions offer powerful ways to tackle oscillatory and circular motion problems. Their derivatives capture these relationships effectively, making them indispensable in calculus and beyond.