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Find the following derivatives. \(w_{s}\) and \(w_{t},\) where \(w=\frac{x-z}{y+z}, x=s+t, y=s t,\) and \(z=s-t\)

Short Answer

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Question: Compute the partial derivatives \(w_s\) and \(w_t\) of the function \(w = \frac{x-z}{y+z}\), given that x, y, and z are functions of s and t. Answer: The partial derivatives \(w_s\) and \(w_t\) of the function are: \(w_s = \frac{-t(x-z)+(x+z)}{(y+z)^2}\) \(w_t = \frac{s(x-z)+(x-z)}{(y+z)^2}\)

Step by step solution

01

Compute the partial derivatives of w with respect to x, y, and z.

Using the quotient rule, which states that \(\frac{d}{dt}\frac{u}{v}=\frac{vu'-u'v}{v^2}\), we can find the partial derivatives of w: \(\frac{\partial w}{\partial x} = \frac{-(y+z)}{(y+z)^2}\) \(\frac{\partial w}{\partial y} = \frac{x-z}{(y+z)^2}\) \(\frac{\partial w}{\partial z} = \frac{-(x+z)}{(y+z)^2}\)
02

Compute the partial derivatives of x, y and z with respect to s and t.

Differentiate x, y, and z with respect to s and t: \(\frac{\partial x}{\partial s} = 1\) \(\frac{\partial x}{\partial t} = 1\) \(\frac{\partial y}{\partial s} = t\) \(\frac{\partial y}{\partial t} = s\) \(\frac{\partial z}{\partial s} = 1\) \(\frac{\partial z}{\partial t} = -1\)
03

Use the chain rule to find the derivatives \(w_s\) and \(w_t\)

To find \(w_s\) and \(w_t\), we use the chain rule: \(w_s = \frac{\partial w}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial s} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial s}\) \(w_t = \frac{\partial w}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial t} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial t}\) Compute \(w_s\) and \(w_t\): \(w_s = \left(-\frac{y+z}{(y+z)^2}\right)(1) + \left(\frac{x-z}{(y+z)^2}\right)(t) + \left(-\frac{x+z}{(y+z)^2}\right)(1)\) \(w_s = \frac{-t(x-z)+(x+z)}{(y+z)^2}\) \(w_t = \left(-\frac{y+z}{(y+z)^2}\right)(1) + \left(\frac{x-z}{(y+z)^2}\right)(s) + \left(-\frac{x+z}{(y+z)^2}\right)(-1)\) \(w_t = \frac{s(x-z)+(x-z)}{(y+z)^2}\)

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Most popular questions from this chapter

Absolute maximum and minimum values Find the absolute maximum and minimum values of the following functions over the given regions \(R\). Use Lagrange multipliers to check for extreme points on the boundary. $$f(x, y)=x^{2}+4 y^{2}+1 ; R=\left\\{(x, y): x^{2}+4 y^{2} \leq 1\right\\}$$

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Let \(w=f(x, y, z)=2 x+3 y+4 z\), which is defined for all \((x, y, z)\) in \(\mathbb{R}^{3}\). Suppose that we are interested in the partial derivative \(w_{x}\) on a subset of \(\mathbb{R}^{3}\), such as the plane \(P\) given by \(z=4 x-2 y .\) The point to be made is that the result is not unique unless we specify which variables are considered independent. a. We could proceed as follows. On the plane \(P\), consider \(x\) and \(y\) as the independent variables, which means \(z\) depends on \(x\) and \(y,\) so we write \(w=f(x, y, z(x, y)) .\) Differentiate with respect to \(x\) holding \(y\) fixed to show that \(\left(\frac{\partial w}{\partial x}\right)_{y}=18,\) where the subscript \(y\) indicates that \(y\) is held fixed. b. Alternatively, on the plane \(P,\) we could consider \(x\) and \(z\) as the independent variables, which means \(y\) depends on \(x\) and \(z,\) so we write \(w=f(x, y(x, z), z)\) and differentiate with respect to \(x\) holding \(z\) fixed. Show that \(\left(\frac{\partial w}{\partial x}\right)_{z}=8,\) where the subscript \(z\) indicates that \(z\) is held fixed. c. Make a sketch of the plane \(z=4 x-2 y\) and interpret the results of parts (a) and (b) geometrically. d. Repeat the arguments of parts (a) and (b) to find \(\left(\frac{\partial w}{\partial y}\right)_{x}\), \(\left(\frac{\partial w}{\partial y}\right)_{z},\left(\frac{\partial w}{\partial z}\right)_{x},\) and \(\left(\frac{\partial w}{\partial z}\right)_{y}\).

a. Show that the point in the plane \(a x+b y+c z=d\) nearest the origin is \(P\left(a d / D^{2}, b d / D^{2}, c d / D^{2}\right),\) where \(D^{2}=a^{2}+b^{2}+c^{2} .\) Conclude that the least distance from the plane to the origin is \(|d| / D\). (Hint: The least distance is along a normal to the plane.) b. Show that the least distance from the point \(P_{0}\left(x_{0}, y_{0}, z_{0}\right)\) to the plane \(a x+b y+c z=d\) is \(\left|a x_{0}+b y_{0}+c z_{0}-d\right| / D\).

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