Problem 11 - The Calculus with Analytic Geometry

Question

In Exercises 11 through 14, find the total derivative \(d u / d t\) by two methods: (a) Use the chain rule; (b) make the substitutions for \(x\) and \(y\) or for \(x, y\), and \(z\) before differentiating. $$ u=y e^{x}+x e^{y} ; x=\cos t ; y=\sin t $$

Step-by-Step Solution

Answer

Using chain rule: du/dt = (y e^x + e^y)(-sin t) + (e^x + x e^y)(cos t)Using substitution: du/dt = e^(cos t)(cos t - sin^2 t) + e^(sin t)(cos^2 t - sin t)

Step 1 - Differentiate using the chain rule

First, express the total derivative of u with respect to t using the chain rule. We have two functions, so: du/dt = (∂u/∂x)(dx/dt) + (∂u/∂y)(dy/dt)We need partial derivatives ∂u/∂x and ∂u/∂y, and derivatives dx/dt and dy/dt.

Step 2 - Compute partial derivatives

Compute the partial derivatives: ∂u/∂x = ∂(y e^x + x e^y)/∂x = y e^x + e^y∂u/∂y = ∂(y e^x + x e^y)/∂y = e^x + x e^y

Step 3 - Compute derivatives of x and y with respect to t

Differentiate x and y with respect to t: dx/dt = d(cos t)/dt = -sin tdy/dt = d(sin t)/dt = cos t

Step 4 - Substitute derivatives into chain rule formula

Substitute the partial and regular derivatives into the chain rule formula: du/dt = (y e^x + e^y)(-sin t) + (e^x + x e^y)(cos t)

Step 5 - Simplify the expression

Expand and combine like terms: du/dt = y e^x (-sin t) + e^y (-sin t) + e^x (cos t) + x e^y (cos t)

Step 6 - Make substitutions and differentiate

Now, substitute x = cos t and y = sin t into the original function u before differentiating: u = (sin t)e^(cos t) + (cos t)e^(sin t)Differentiate this function directly with respect to t: du/dt = d((sin t)e^(cos t) + (cos t)e^(sin t))/dt

Step 7 - Differentiate term-by-term

Differentiate each term separately using the product rule: For (sin t)e^(cos t):d/dt((sin t)e^(cos t)) = (cos t)e^(cos t) + (sin t)(-sin t)e^(cos t)For (cos t)e^(sin t):d/dt((cos t)e^(sin t)) = (-sin t)e^(sin t) + (cos t)(cos t)e^(sin t)

Step 8 - Combine the derivatives

Add the derivatives from Step 7: du/dt = (cos t)e^(cos t) + (sin t)(-sin t)e^(cos t) + (-sin t)e^(sin t) + (cos t)(cos t)e^(sin t) = e^(cos t)(cos t - sin^2 t) + e^(sin t)(cos^2 t - sin t)

Key Concepts

chain rule

The chain rule helps us differentiate functions composed of other functions. Imagine you have a function of two variables, such as \( u = y e^{x} + x e^{y} \), and each variable is a function of another variable, like \( x = \cos t \) and \( y = \sin t \). The chain rule allows us to find the total derivative of \( u \) with respect to \( t \).
Start by writing \( du/dt \) as:
\( du / dt = (\frac{\partial u}{\partial x})(\frac{dx}{dt}) + (\frac{\partial u}{\partial y})(\frac{dy}{dt}) \)
Here, we need the partial derivatives \( \frac{\partial u}{\partial x} \) and \( \frac{\partial u}{\partial y} \), and the regular derivatives \( dx / dt \) and \( dy / dt \).
Plug these values back into the formula to get the total derivative.

partial derivatives

Partial derivatives tell us how a multivariable function changes as one variable changes while keeping others fixed. For \( u = y e^{x} + x e^{y} \), we need to find:
- \( \frac{\partial u}{\partial x} \)
- \( \frac{\partial u}{\partial y} \)
To compute \( \frac{\partial u}{\partial x} \), treat \( y \) as constant and differentiate with respect to \( x \): \( \frac{\partial }{\partial x} (y e^{x} + x e^{y}) = y e^{x} + e^{y} \)
To compute \( \frac{\partial u}{\partial y} \), treat \( x \) as constant and differentiate with respect to \( y \): \( \frac{\partial u}{\partial y} = e^{x} + x e^{y} \).
These partial derivatives feed into our chain rule calculation.

product rule

The product rule is used to differentiate products of two functions. For functions \( f(t) \) and \( g(t) \), the derivative is:
\( (fg)^{'} = f^{'}g + fg^{'} \)
We use this rule in our problem where each term in \( u = (\sin t)e^{\cos t} + (\cos t)e^{\sin t} \) is a product:
For \( (\sin t)e^{\cos t} \), apply the product rule:
\( d/dt(\sin t)e^{\cos t}) = (\cos t)e^{\cos t} + (\sin t)(-\sin t)e^{\cos t} \)
For \( (\cos t)e^{\sin t} \):
\( d/dt(\cos t)e^{\sin t}) = (-\sin t)e^{\sin t} + (\cos t)(\cos t)e^{\sin t} \)
Combine all these results to get the final derivative.

differentiation techniques

Differentiation techniques help us find the rate at which one quantity changes relative to another. Here, we used:
- The chain rule to connect variables through intermediate functions.
- Partial derivatives to capture how multi-variable functions change.
- The product rule for differentiating products of functions.
These techniques are the backbone of solving problems involving multi-variable functions and their compositions. Practice them in various scenarios to get comfortable.

trigonometric functions

Trigonometric functions like \( \sin t \) and \( \cos t \) have specific differentiation rules:
- \( d/dt (\sin t) = \cos t \)
- \( d/dt (\cos t) = -\sin t \)
These derivatives are essential in our example, where \( x = \cos t \) and \( y = \sin t \).
When making substitutions and differentiating, always appeal to these basic rules. For example, to find \( dy/dt \) where \( y = \sin t \), we get \( \cos t \). Similarly, for \( dx/dt \) where \( x = \cos t \), the result is \( -\sin t \). These are straightforward but crucial steps in solving the overall problem.