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Find \(\partial z / \partial u\) and \(\partial z / \partial v .\) The variables are restricted to domains on which the functions are defined. $$z=\tan ^{-1}(x / y), x=u^{2}+v^{2}, y=u^{2}-v^{2}$$

Short Answer

Expert verified
\(\frac{\partial z}{\partial u} = \frac{-4uv^2}{(1+t^2)(u^2-v^2)^2}\), \(\frac{\partial z}{\partial v} = \frac{4vu^2}{(1+t^2)(u^2-v^2)^2}\).

Step by step solution

01

Express z in terms of u and v

Start by substituting the expressions for \( x \) and \( y \) in the formula for \( z \):\[ z = \tan^{-1} \left( \frac{x}{y} \right) = \tan^{-1} \left( \frac{u^2 + v^2}{u^2 - v^2} \right) \]Now, \( z \) is expressed in terms of \( u \) and \( v \).
02

Compute the partial derivative \( \frac{\partial z}{\partial u} \)

Using the chain rule and the derivative of \( \tan^{-1}(t) \), which is \( \frac{1}{1+t^2} \), we find:\[ \frac{\partial z}{\partial u} = \frac{1}{1 + \left( \frac{u^2 + v^2}{u^2 - v^2} \right)^2} \cdot \frac{\partial}{\partial u} \left( \frac{u^2 + v^2}{u^2 - v^2} \right) \]Calculate the derivative of the quotient:\[ \frac{\partial}{\partial u} \left( \frac{u^2 + v^2}{u^2 - v^2} \right) = \frac{(2u)(u^2 - v^2) - (u^2 + v^2)(2u)}{(u^2 - v^2)^2} = \frac{-4uv^2}{(u^2 - v^2)^2} \]Thus, \[ \frac{\partial z}{\partial u} = \frac{-4uv^2}{(1 + t^2)(u^2 - v^2)^2} \]where \( t = \frac{u^2 + v^2}{u^2 - v^2} \).
03

Compute the partial derivative \( \frac{\partial z}{\partial v} \)

Similarly, we use the chain rule for the derivative \( \tan^{-1}(t) \):\[ \frac{\partial z}{\partial v} = \frac{1}{1 + \left( \frac{u^2 + v^2}{u^2 - v^2} \right)^2} \cdot \frac{\partial}{\partial v} \left( \frac{u^2 + v^2}{u^2 - v^2} \right) \]Calculate the derivative of the quotient:\[ \frac{\partial}{\partial v} \left( \frac{u^2 + v^2}{u^2 - v^2} \right) = \frac{(2v)(u^2 - v^2) - (u^2 + v^2)(-2v)}{(u^2 - v^2)^2} = \frac{4vu^2}{(u^2 - v^2)^2} \]Thus, \[ \frac{\partial z}{\partial v} = \frac{4vu^2}{(1 + t^2)(u^2 - v^2)^2} \]where \( t = \frac{u^2 + v^2}{u^2 - v^2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule in Partial Differentiation
When dealing with partial differentiation, the **chain rule** is a fundamental tool. It helps us differentiate functions that are nested or composed of terms dependent on each other. Imagine a function like \( z = \tan^{-1}(x / y) \), where both \( x \) and \( y \) are themselves functions of other variables like \( u \) and \( v \). To find the derivative of \( z \) with respect to \( u \) or \( v \), we must consider how changes in \( u \) and \( v \) affect \( x \) and \( y \), and thus affect \( z \).

Start by expressing \( z \) in terms of \( u \) and \( v \) directly. Then, use the chain rule to differentiate. The **main idea** is to multiply the derivative of the outer function by the derivative of the inner function. So, for \( \frac{\partial z}{\partial u} \), you'd compute the derivative of \( \tan^{-1} \) with respect to its input and multiply that by the derivative of the fraction inside with respect to \( u \). You repeat similarly for \( \frac{\partial z}{\partial v} \).

Overall, the chain rule in multivariable calculus allows us to handle more complex relationships between variables smoothly.
Inverse Trigonometric Functions
The world of inverse trigonometric functions, such as **\( \tan^{-1} \)**, adds fascinating layers to calculus. These functions allow us to reverse trigonometric operations, revealing angles rather than ratios. In the context of differentiation, knowledge about their derivatives is crucial.

For \( \tan^{-1}(t) \), the derivative is \[ \frac{1}{1+t^2} \]. This formula is key when we're differentiating expressions like \( \tan^{-1}(\frac{x}{y}) \). The formula shows how any change in \( t \), our ratio, will scale the change in angle, \( z \). It is crucial when applied using the chain rule, as it captures how the angle changes with respect to the root variables \( u \) and \( v \).

These functions appear often in multivariable calculus, where relationships between variables are captured in ratios. Differentiating them correctly ensures accurate representations of these relationships, helping us understand how scenarios change with slight variable adjustments.
Multivariable Calculus and Partial Derivatives
In **multivariable calculus**, we extend the concepts of calculus to functions of multiple variables. This branch deals with situations where there is more than one input affecting an output. Partial derivatives are a key concept here.

A partial derivative measures how a function changes as one of the variables changes, keeping others constant. For a function \( z = f(u, v) \), \( \frac{\partial z}{\partial u} \) shows the rate of change of \( z \) with respect to \( u \), assuming \( v \) is held steady. Similarly, \( \frac{\partial z}{\partial v} \) reflects changes with \( v \). This is crucial when studying systems in physics, engineering or economics, where multiple factors can fluctuate independently.

Multivariable calculus enables us to deal with real-world problems that can't be captured with single-variable approaches. By using partial derivatives and the chain rule together, we gain insight into how each variable influences results in complex scenarios. This knowledge is powerful in optimizing systems, predicting changes, and understanding interconnected variables.

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Most popular questions from this chapter

The tastiness, \(T\), of a soup depends on the volume, \(V\) of the soup in the pot and the quantity, \(S,\) of salt in the soup. If you have more soup, you need more salt to make it taste good. Match the three stories (a)-(c) to the three statements (I)-(III) about partial derivatives. (a) I started adding salt to the soup in the pot. At first the taste improved, but eventually the soup became too salty and continuing to add more salt made it worse. (b) The soup was too salty, so I started adding unsalted soup. This improved the taste at first, but eventually there was too much soup for the salt, and continuing to add unsalted soup just made it worse. (c) The soup was too salty, so adding more salt would have made it taste worse. I added a quart of unsalted soup instead. Now it is not salty enough, but I can improve the taste by adding salt. (I) \(\partial^{2} T / \partial V^{2} < 0\) (II) \(\partial^{2} T / \partial S^{2} < 0\) (III) \(\partial^{2} T / \partial V \partial S > 0\)

True or false? Give reasons for your answer. If you know the gradient vector of \(f\) at \((a, b)\) then you can find the directional derivative \(f_{\vec{u}}(a, b)\) for any unit vector \(\vec{u}\)

(a) Give \(Q\), the point at a distance of 0.1 from \(P=\) (4,5) in the direction of \(\vec{v}=-\vec{i}+3 \vec{j} .\) Give five decimal places in your answer. (b) Use \(P\) and \(Q\) to approximate the directional derivative of \(f(x, y)=\sqrt{x+y}\) in the direction of \(\vec{v}\) (c) Give the exact value for the directional derivative you estimated in part (b).

(a) Find the differential of \(g(u, v)=u^{2}+u v\) (b) Use your answer to part (a) to estimate the change in \(g\) as you move from (1,2) to (1.2,2.1)

Find the gradient at the point. $$f(x, y)=x^{2} y+7 x y^{3}, \text { at }(1,2)$$

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