Problem 4 - The Calculus with Analytic Geometry

Question

In Exercises 1 through 4 , find the indicated partial derivative by two methods: (a) Use the chain rule; (b) make the substitutions for \(x\) and \(y\) before differentiating. $$ u=x^{2}+y^{2} ; x=\cosh r \cos t ; y=\sinh r \sin t ; \frac{\partial u}{\partial r} ; \frac{\partial u}{\partial t} $$

Step-by-Step Solution

Answer

\( \frac{\partial u}{\partial r} = 2 \cosh r \sinh r \); \( \frac{\partial u}{\partial t} = 2 \cos t \sin t (\sinh^2 r - \cosh^2 r) \).

Step 1: Write Given Functions

Given, functions are: \( u = x^2 + y^2 \), \( x = \cosh r \cos t \), and \( y = \sinh r \sin t \). We need to find \( \frac{\partial u}{\partial r} \) and \( \frac{\partial u}{\partial t} \).

Step 2: Calculate \( \frac{\partial u}{\partial r} \) using Chain Rule

First, find the partial derivatives of \( u \) with respect to \( x \) and \( y \): \( \frac{\partial u}{\partial x} = 2x \) and \( \frac{\partial u}{\partial y} = 2y \). Next, find the partial derivatives of \( x \) and \( y \) with respect to \( r \): \( \frac{\partial x}{\partial r} = \sinh r \cos t \) and \( \frac{\partial y}{\partial r} = \cosh r \sin t \). Now apply the chain rule: \( \frac{\partial u}{\partial r} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial r} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial r} \) Substituting the values, we get: \(\frac{\partial u}{\partial r} = 2x (\sinh r \cos t) + 2y (\cosh r \sin t) \).

Step 3: Calculate \( \frac{\partial u}{\partial t} \) using Chain Rule

First, find the partial derivatives of \( x \) and \( y \) with respect to \( t \): \( \frac{\partial x}{\partial t} = -\cosh r \sin t \) and \( \frac{\partial y}{\partial t} = \sinh r \cos t \). Now applying the chain rule: \( \frac{\partial u}{\partial t} = \frac{\partial u}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial u}{\partial y} \frac{\partial y}{\partial t} \) Substituting the values, we get: \(\frac{\partial u}{\partial t} = 2x (-\cosh r \sin t) + 2y (\sinh r \cos t)\).

Step 4: Make Substitutions before Differentiation

Substitute \( x = \cosh r \cos t \) and \( y = \sinh r \sin t \) into \( u = x^2 + y^2 \): \( u = (\cosh r \cos t)^2 + (\sinh r \sin t)^2 \) Simplify: \( u = \cosh^2 r \cos^2 t + \sinh^2 r \sin^2 t \). Now, find the partial derivatives \( \frac{\partial u}{\partial r} \) and \( \frac{\partial u}{\partial t} \).

Step 5: Differentiate \( u \) with respect to \( r \)

Differentiating \( u = \cosh^2 r \cos^2 t + \sinh^2 r \sin^2 t \) with respect to \( r \): \( \frac{\partial u}{\partial r} = 2 \cosh r \sinh r \cos^2 t + 2 \sinh r \cosh r \sin^2 t \) Simplify: \( \frac{\partial u}{\partial r} = 2 \cosh r \sinh r (\cos^2 t + \sin^2 t) \) Finally, we know \( \cos^2 t + \sin^2 t = 1 \): \( \frac{\partial u}{\partial r} = 2 \cosh r \sinh r \).

Step 6: Differentiate \( u \) with respect to \( t \)

Differentiating \( u = \cosh^2 r \cos^2 t + \sinh^2 r \sin^2 t \) with respect to \( t \): \( \frac{\partial u}{\partial t} = -2 \cosh^2 r \cos t \sin t + 2 \sinh^2 r \cos t \sin t \) Simplify: \( \frac{\partial u}{\partial t} = 2 \cos t \sin t (\sinh^2 r - \cosh^2 r) \).

Key Concepts

chain rule

The chain rule is fundamental for computing derivatives of composite functions. When you have a function composed of other functions, you need to consider how changes in one variable affect the entire system. In the given problem, we use the chain rule to find the partial derivatives of the function \( u \) with respect to \( r \) and \( t \). Essentially, the chain rule states that for a function \( u = f(x, y) \), where \( x = g(r, t) \) and \( y = h(r, t) \): \[ \frac{\text{d} u}{\text{d} r} = \frac{\text{d} u}{\text{d} x} \frac{\text{d} x}{\text{d} r} + \frac{\text{d} u}{\text{d} y} \frac{\text{d} y}{\text{d} r} \ \frac{\text{d} u}{\text{d} t} = \frac{\text{d} u}{\text{d} x} \frac{\text{d} x}{\text{d} t} + \frac{\text{d} u}{\text{d} y} \frac{\text{d} y}{\text{d} t} \ \] By applying these formulas, we see how the change in \( r \) or \( t \) affects the entire function \( u \). This method ensures all variables are properly accounted for.

differentiation

Differentiation involves finding the derivative of a function, which indicates the rate of change. In this problem, we need partial derivatives of \( u = x^2 + y^2 \). A partial derivative is the derivative with respect to one variable while holding others constant. Differentiating \( u \) with respect to \( x \) and \( y \): \[ \frac{\text{d} u}{\text{d} x} = 2x \quad \text{and} \quad \frac{\text{d} u}{\text{d} y} = 2y \ \] Next, differentiate \( x = \text{cosh} \thinspace r \text{cos} \thinspace t \) and \( y = \text{sinh} \thinspace r \text{sin} \thinspace t \) with respect to \( r \): \[ \frac{\text{d} x}{\text{d} r} = \text{sinh} \thinspace r \text{cos} \thinspace t \quad \text{and} \quad \frac{\text{d} y}{\text{d} r} = \text{cosh} \thinspace r \text{sin} \thinspace t \ \] This step is crucial in our calculations because it provides the intermediate results needed for the chain rule application.

substitution method

The substitution method involves directly replacing the variables with their expressions before differentiating. Instead of using the chain rule, we convert \( u \), \( x \), and \( y \) into functions of \( r \) and \( t \), then differentiate. For example, by substituting: \[ u = (\text{cosh} \thinspace r \text{cos} \thinspace t)^2 + (\text{sinh} \thinspace r \text{sin} \thinspace t)^2 \ \] Then simplify: \[ u = \text{cosh}^2 \thinspace r \thinspace \text{cos}^2 \thinspace t + \text{sinh}^2 \thinspace r \thinspace \text{sin}^2 \thinspace t \ \] After simplifying, we efficiently find the partial derivatives \( \frac{\text{d} u}{\text{d} r} \) and \( \frac{\text{d} u}{\text{d} t} \) by treating \( u \) as a function directly of \( r \) and \( t \).

cosh and sinh functions

The hyperbolic functions cosh and sinh behave similarly to the trigonometric functions cosine and sine, but are based on hyperbolas instead of circles. In the given exercise, \( \text{cosh} \thinspace r \) and \( \text{sinh} \thinspace r \) are used for coordinates that describe hyperbolic angles. Here are the definitions: \[ \text{cosh} \thinspace r = \frac{e^r + e^{-r}}{2} \quad \text{and} \quad \text{sinh} \thinspace r = \frac{e^r - e^{-r}}{2} \ \] These identities help when differentiating. For instance: \[ \frac{\text{d}}{\text{d} r} (\text{cosh} \thinspace r) = \text{sinh} \thinspace r \quad and \quadr \frac{\text{d}}{\text{d} r} (\text{sinh} \thinspace r) = \text{cosh} \thinspace r \ \] Understanding these functions’ properties makes solving such problems more straightforward. They also simplify calculations involving hyperbolic coordinates.