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Chapter 15: Problem 30

Partial derivatives Find the first partial derivatives of the following functions. $$h(u, v)=\sqrt{\frac{u v}{u-v}}$$

Short Answer

Expert verified
Question: Find the first partial derivatives of the function h(u, v) = 鈭�(uv / (u - v)) with respect to u and v. Answer: The first partial derivatives of the function h(u, v) are: $$\frac{\partial h}{\partial u} = \frac{1}{2}\left(\frac{u v}{u - v}\right)^{-\frac{1}{2}} \left(\frac{v(u-v) - uv}{(u-v)^2}\right)$$ $$\frac{\partial h}{\partial v} = \frac{1}{2}\left(\frac{u v}{u - v}\right)^{-\frac{1}{2}} \left(\frac{u(u - v) + u^2}{(u-v)^2}\right)$$

Step by step solution

01

Find the partial derivative with respect to u

To find the partial derivative of h with respect to u, we'll first differentiate the outer function (the square root) and then differentiate the inner function (the fraction) using the chain rule. First, let's rewrite the function as: $$h(u, v) = \left(\frac{u v}{u - v}\right)^{\frac{1}{2}}$$ Now, let's find \(\frac{\partial h}{\partial u}\): $$\frac{\partial h}{\partial u} = \frac{1}{2} \left(\frac{u v}{u - v}\right)^{-\frac{1}{2}} \frac{\partial}{\partial u}\left(\frac{u v}{u - v}\right)$$ We'll now use the quotient rule to differentiate the inner fraction: $$\frac{\partial}{\partial u}\left(\frac{u v}{u - v}\right) = \frac{v(u-v) - uv}{(u-v)^2}$$ So, the partial derivative of h with respect to u is: $$\frac{\partial h}{\partial u} = \frac{1}{2}\left(\frac{u v}{u - v}\right)^{-\frac{1}{2}} \left(\frac{v(u-v) - uv}{(u-v)^2}\right)$$
02

Find the partial derivative with respect to v

Similarly, we'll find the partial derivative of h with respect to v: $$\frac{\partial h}{\partial v} = \frac{1}{2} \left(\frac{u v}{u - v}\right)^{-\frac{1}{2}} \frac{\partial}{\partial v}\left(\frac{u v}{u - v}\right)$$ We'll again use the quotient rule to differentiate the inner fraction: $$\frac{\partial}{\partial v}\left(\frac{u v}{u - v}\right) = \frac{u(u - v) + u^2}{(u-v)^2}$$ So, the partial derivative of h with respect to v is: $$\frac{\partial h}{\partial v} = \frac{1}{2}\left(\frac{u v}{u - v}\right)^{-\frac{1}{2}} \left(\frac{u(u - v) + u^2}{(u-v)^2}\right)$$ In conclusion, the first partial derivatives of the function h(u, v) are: $$\frac{\partial h}{\partial u} = \frac{1}{2}\left(\frac{u v}{u - v}\right)^{-\frac{1}{2}} \left(\frac{v(u-v) - uv}{(u-v)^2}\right)$$ $$\frac{\partial h}{\partial v} = \frac{1}{2}\left(\frac{u v}{u - v}\right)^{-\frac{1}{2}} \left(\frac{u(u - v) + u^2}{(u-v)^2}\right)$$

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Key Concepts

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

Chain Rule
Understanding the chain rule is fundamental when dealing with functions within functions, a common occurrence in calculus. Imagine we have a function tucked inside another function, much like a nesting doll. When we differentiate such composite functions, we peel back the layers one by one. This is exactly what the chain rule allows us to do.

Let's take a closer look at our given function, where we have a square root (the outer function) and a fraction inside it (the inner function). Applying the chain rule, we differentiate the outer function, keeping the inner one intact, then multiply by the derivative of the inner function. If our inner function is complex, involving other rules of differentiation, such as the quotient rule, then we apply those as necessary to find the derivative of the inner function. The chain rule is an indispensable tool, especially in multivariable calculus, where functions become more intricate and interwoven.
Quotient Rule
One of the trickier scenarios encountered in calculus is taking the derivative of a fraction, where both the numerator and the denominator are functions themselves鈥攈ence the quotient rule comes to the rescue. It states that the derivative of a fraction is the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator.

The quotient rule plays a pivotal role in our exercise. To find the partial derivatives of our function with respect to each variable, we needed to apply the quotient rule to differentiate the fraction \(\frac{uv}{u-v}\). This is because we have one function (the product of \(u\) and \(v\)) divided by another (the difference of \(u\) and \(v\)). It's essentially about finding the rate of change of the ratio between two changing quantities.
Multivariable Calculus
Multivariable calculus expands the field of calculus to functions with more than one variable, like the function in our exercise \(h(u, v)\). Instead of dealing with just one independent variable, we consider multiple inputs and how each input independently alters the outcome. In our example, we find the partial derivatives with respect to \(u\) and \(v\), effectively determining how \(h\) changes as \(u\) varies while holding \(v\) constant, and vice versa.

This concept is crucial because it reflects the complexity of real-world scenarios where multiple factors can influence a result. By using partial derivatives, we can analyze the impact of each variable in isolation, a core aspect of modeling and solving problems in physics, engineering, economics, and many other fields. Furthermore, mastering the application of the chain and quotient rules within multivariable calculus is a testament to the interconnected nature of mathematical principles.

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Most popular questions from this chapter

Second Derivative Test Suppose the conditions of the Second Derivative Test are satisfied on an open disk containing the point \((a, b) .\) Use the test to prove that if \((a, b)\) is a critical point of \(f\) at which \(f_{x}(a, b)=f_{y}(a, b)=0\) and \(f_{x x}(a, b) < 0 < f_{\text {vy }}(a, b)\) or \(f_{y y}(a, b) < 0 < f_{x x}(a, b),\) then \(f\) has a saddle point at \((a, b)\)

A snapshot (frozen in time) of a set of water waves is described by the function \(z=1+\sin (x-y),\) where \(z\) gives the height of the waves and \((x, y)\) are coordinates in the horizontal plane \(z=0\) a. Use a graphing utility to graph \(z=1+\sin (x-y)\) b. The crests and the troughs of the waves are aligned in the direction in which the height function has zero change. Find the direction in which the crests and troughs are aligned. c. If you were surfing on one of these waves and wanted the steepest descent from the crest to the trough, in which direction would you point your surfboard (given in terms of a unit vector in the \(x y\) -plane)? d. Check that your answers to parts (b) and (c) are consistent with the graph of part (a).

Extreme points on flattened spheres The equation \(x^{2 n}+y^{2 n}+z^{2 n}=1,\) where \(n\) is a positive integer, describes a flattened sphere. Define the extreme points to be the points on the flattened sphere with a maximum distance from the origin. a. Find all the extreme points on the flattened sphere with \(n=2 .\) What is the distance between the extreme points and the origin? b. Find all the extreme points on the flattened sphere for integers \(n>2 .\) What is the distance between the extreme points and the origin? c. Give the location of the extreme points in the limit as \(n \rightarrow \infty\) What is the limiting distance between the extreme points and the origin as \(n \rightarrow \infty ?\)

Suppose you make monthly deposits of \(P\) dollars into an account that earns interest at a monthly rate of \(p \% .\) The balance in the account after \(t\) years is \(B(P, r, t)=P\left(\frac{(1+r)^{12 t}-1}{r}\right),\) where \(r=\frac{p}{100}\) (for example, if the annual interest rate is \(9 \%,\) then \(p=\frac{9}{12}=0.75\) and \(r=0.0075) .\) Let the time of investment be fixed at \(t=20\) years. a. With a target balance of \(\$ 20,000,\) find the set of all points \((P, r)\) that satisfy \(B=20,000 .\) This curve gives all deposits \(P\) and monthly interest rates \(r\) that result in a balance of \(\$ 20,000\) after 20 years. b. Repeat part (a) with \(B=\$ 5000, \$ 10,000, \$ 15,000,\) and \(\$ 25,000,\) and draw the resulting level curves of the balance function.

The electric potential function for two positive charges, one at (0,1) with twice the strength of the charge at \((0,-1),\) is given by $$\varphi(x, y)=\frac{2}{\sqrt{x^{2}+(y-1)^{2}}}+\frac{1}{\sqrt{x^{2}+(y+1)^{2}}}$$ a. Graph the electric potential using the window $$[-5,5] \times[-5,5] \times[0,10]$$ b. For what values of \(x\) and \(y\) is the potential \(\varphi\) defined? c. Is the electric potential greater at (3,2) or (2,3)\(?\) d. Describe how the electric potential varies along the line \(y=x\)
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