Problem 6 - The Calculus with Analytic Geometry

Question

In Exercises 5 through 10, find the indicated partial derivative by using the chain rule. $$ u=x e^{-y} ; x=\tan ^{-1}(r s t) ; y=\ln (3 r s+5 s t) ; \frac{\partial u}{\partial r} ; \frac{\partial u}{\partial s} ; \frac{\partial u}{\partial t} $$

Step-by-Step Solution

Answer

The partial derivatives are as follows: \(\frac{\rho u}{\rho r} = e^{-y} \frac{s t}{1 + (r s t)^2} - x e^{-y} \frac{3 s}{3 r s + 5 s t}\), \(\frac{\rho u}{\rho s} = e^{-y} \frac{r t}{1 + (r s t)^2} - x e^{-y} \frac{3 r + 5 t}{3 r s + 5 s t}\), and \(\frac{\rho u}{\rho t} = e^{-y} \frac{r s}{1 + (r s t)^2} - x e^{-y} \frac{5 s}{3 r s + 5 s t}\).

Step 1 - Identify given functions

The functions provided are: \[ u = x e^{-y} \]\[ x = \tan^{-1}(r s t) \]\[ y = \ln(3 r s + 5 s t) \]We need to find the partial derivatives: \(\frac{\frac{\rho u}{\rho r}} \), \(\frac{\frac{\rho u}{\rho s}} \), and \(\frac{\frac{\rho u}{\rho t}} \) using the chain rule.

Step 2 - Compute \(\frac{\rho u}{\rho x}\) and \(\frac{\rho u}{\rho y}\)

First, compute the partial derivatives of \(u\) with respect to \(x\) and \(y\): \[\frac{\rho u}{\rho x} = e^{-y}\]\[\frac{\rho u}{\rho y} = -x e^{-y}\]

Step 3 - Compute \(\frac{\rho x}{\rho r}\) \(\frac{\rho x}{\rho s}\) and \(\frac{\rho x}{\rho t}\)

Next, compute the partial derivatives of \(x\) with respect to \(r\), \(s\), and \(t\):\[\frac{\rho x}{\rho r} = \frac{s t}{1 + (r s t)^2}\]\[\frac{\rho x}{\rho s} = \frac{r t}{1 + (r s t)^2}\]\[\frac{\rho x}{\rho t} = \frac{r s}{1 + (r s t)^2}\]

Step 4 - Compute \(\frac{\rho y}{\rho r}\) \(\frac{\rho y}{\rho s}\) and \(\frac{\rho y}{\rho t}\)

Then, compute the partial derivatives of \(y\) with respect to \(r\), \(s\), and \(t\):\[\frac{\rho y}{\rho r} = \frac{3 s}{3 r s + 5 s t}\]\[\frac{\rho y}{\rho s} = \frac{3 r + 5 t}{3 r s + 5 s t}\]\[\frac{\rho y}{\rho t} = \frac{5 s}{3 r s + 5 s t}\]

Step 5 - Apply chain rule for \(\frac{\rho u}{\rho r}\)

Apply the chain rule to find \(\frac{\rho u}{\rho r}\):\[\frac{\rho u}{\rho r} = \frac{\rho u}{\rho x} \frac{\rho x}{\rho r} + \frac{\rho u}{\rho y} \frac{\rho y}{\rho r}\]Substitute the partial derivatives found in previous steps:\[\frac{\rho u}{\rho r} = e^{-y} \frac{s t}{1 + (r s t)^2} + (-x e^{-y}) \frac{3 s}{3 r s + 5 s t}\]

Step 6 - Apply chain rule for \(\frac{\rho u}{\rho s}\)

Apply the chain rule to find \(\frac{\rho u}{\rho s}\):\[\frac{\rho u}{\rho s} = \frac{\rho u}{\rho x} \frac{\rho x}{\rho s} + \frac{\rho u}{\rho y} \frac{\rho y}{\rho s}\]Substitute the partial derivatives found in previous steps:\[\frac{\rho u}{\rho s} = e^{-y} \frac{r t}{1 + (r s t)^2} + (-x e^{-y}) \frac{3 r + 5 t}{3 r s + 5 s t}\]

Step 7 - Apply chain rule for \(\frac{\rho u}{\rho t}\)

Apply the chain rule to find \(\frac{\rho u}{\rho t}\):\[\frac{\rho u}{\rho t} = \frac{\rho u}{\rho x} \frac{\rho x}{\rho t} + \frac{\rho u}{\rho y} \frac{\rho y}{\rho t}\]Substitute the partial derivatives found in previous steps:\[\frac{\rho u}{\rho t} = e^{-y} \frac{r s}{1 + (r s t)^2} + (-x e^{-y}) \frac{5 s}{3 r s + 5 s t}\]

Key Concepts

Calculus

Calculus helps us understand changes. It is essential for many fields such as physics, engineering, economics, statistics, and so much more. It consists of two main branches: differential calculus and integral calculus.

Differential calculus focuses on the concept of the derivative, which measures how a function changes as its input changes. This branch answers questions like: 'How fast is a car going at a specific moment?'.

Integral calculus, on the other hand, deals with accumulation of quantities, such as areas under curves. It answers questions like: 'How far has a car traveled over a period of time?' by integrating the speed function.

In our problem, we're dealing with partial derivatives, which comes under the wing of differential calculus. We use the chain rule to compute these derivatives when functions are nested within other functions. Understanding the chain rule is crucial in solving complex calculus problems efficiently.

Partial Derivatives

Partial derivatives are a type of derivative where we change one variable while keeping the others constant. This is very useful when dealing with functions of multiple variables.

For instance, if you have a function \(z = f(x, y)\), the partial derivative with respect to \(x\) is written as \(\frac{\rho z}{\rho x}\), and it shows how \(z\) changes as \(x\) changes, keeping \(y\) constant.

In the given exercise, we have a more complex scenario. The function \(u\) is dependent on \(x\) and \(y\). Additionally, \(x\) and \(y\) are functions of \(r, s,\) and \(t\). Thus, to find \(\frac{\rho u}{\rho r}\), \(\frac{\rho u}{\rho s}\), and \(\frac{\rho u}{\rho t}\), we need to consider how \(u\) changes as each of \(r, s,\) and \(t\) change.

Chain Rule

The chain rule is a fundamental tool in calculus for finding the derivative of a composite function. It’s like a recipe that tells us how to take the derivative of nested functions.

In its simplest form, if you have a function \(y = f(g(x))\), then the derivative of \(y\) with respect to \(x\) using the chain rule is \(\frac{dy}{dx} = f'(g(x)) \times g'(x)\).

For partial derivatives, the chain rule is extended. For example, in our specific exercise, we have \(u = u(x, y)\) where \(x\) and \(y\) are functions of \(r, s,\) and \(t\). To find the partial derivative of \(u\) with respect to \(r\), we use:

\[ \frac{\rho u}{\rho r} = \frac{\rho u}{\rho x} \frac{\rho x}{\rho r} + \frac{\rho u}{\rho y} \frac{\rho y}{\rho r} \]

This rule helps decompose the derivative into simpler parts. By first calculating the intermediate partial derivatives (steps 2, 3, and 4 in our solution), we can then piece them together using the chain rule (steps 5, 6, and 7) to get the final partial derivatives \(\frac{\rho u}{\rho r}\), \(\frac{\rho u}{\rho s}\), and \(\frac{\rho u}{\rho t}\).