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Consider the function \(f(x, y, z)=x y+y z^{2}+x z^{3}\). Find the gradient of \(f\) : ( ____________, ____________) Find the gradient of \(f\) at the point (2,3,5) . ( ____________, ____________) Find the rate of change of the function \(f\) at the point (2,3,5) in the direction \(\mathbf{u}=\langle 4 / \sqrt{66},-5 / \sqrt{66},-5 / \sqrt{66}\rangle .\)

Short Answer

Expert verified
The gradient of f is \(\nabla f = (y+z^3, x+z^2, 2yz+3xz^2)\). At the point (2, 3, 5), the gradient is (128, 27, 160). The rate of change of the function at this point in the given direction is \(\frac{-423}{\sqrt{66}}\).

Step by step solution

01

Write down the given function

We have a function in three variables: \( f(x, y, z) = xy + yz^2 + xz^3 \).
02

Find the gradient of the function

The gradient of a function is a vector containing the partial derivatives with respect to each variable. Calculate the partial derivatives: \(\frac{\partial f}{\partial x} = y + z^3\) \(\frac{\partial f}{\partial y} = x + z^2\) \(\frac{\partial f}{\partial z} = 2yz + 3xz^2\) The gradient of function f is therefore: \(\nabla f = (y+z^3, x+z^2, 2yz+3xz^2)\)
03

Find the gradient at the point (2, 3, 5)

Substitute the values of the point (2, 3, 5) into the gradient to find its value at this specific point: \(\nabla f(2, 3, 5) = (3+5^3, 2+5^2, 2(3)(5)+3(2)(5^2)) = (128, 27, 160)\) Thus, the gradient of f at the point (2, 3, 5) is (128, 27, 160).
04

Find the rate of change in the given direction

The rate of change of f in the direction of the vector \(\mathbf{u} = \langle 4/\sqrt{66}, -5/\sqrt{66}, -5/\sqrt{66} \rangle\) can be found by taking the dot product of the gradient and the direction vector: \(Rate\:of\:change = \nabla f \cdot \mathbf{u}\) \(= (128, 27, 160) \cdot \left(\frac{4}{\sqrt{66}},\frac{-5}{\sqrt{66}},\frac{-5}{\sqrt{66}}\right)\) \(= \frac{128 \cdot 4 - 27 \cdot 5 - 160 \cdot 5}{\sqrt{66}}\) \(= \frac{512 - 135 - 800}{\sqrt{66}}\) \(= \frac{-423}{\sqrt{66}}\) The rate of change of the function f at the point (2, 3, 5) in the direction of vector \(\mathbf{u}\) is \(\frac{-423}{\sqrt{66}}\).

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Key Concepts

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

Partial Derivatives
Partial derivatives are taken for functions with multiple variables. Instead of differentiating the whole function with respect to one variable while ignoring others, they allow us to focus on the effect of change in one variable at a time. In our exercise, the function \(f(x, y, z) = xy + yz^2 + xz^3\) comprises three variables: \(x\), \(y\), and \(z\). The partial derivative of \(f\) with respect to \(x\) is found by treating \(y\) and \(z\) as constants. This yields \(\frac{\partial f}{\partial x} = y + z^3\). Likewise, \(\frac{\partial f}{\partial y} = x + z^2\) and \(\frac{\partial f}{\partial z} = 2yz + 3xz^2\) are calculated similarly. Partial derivatives are helpful to understand how \(f\) changes as each individual variable changes, keeping others fixed.
Rate of Change
The concept of rate of change is vital in understanding how a quantity shifts over time or concerning another variable. In multivariable calculus, it's not just about changing in time, but also in spatial dimensions. One way to measure this is through the gradient, as it provides the direction and magnitude of the steepest ascent. Specifically, in our example, we looked for the rate of change in a specific direction given by the vector \(\mathbf{u}\). The formula involves computing the dot product of the gradient (calculated as \(abla f(x, y, z)\)) with the direction vector \(\mathbf{u}\). This resulted in the expression for the rate of change: \(\frac{-423}{\sqrt{66}}\). It tells us how fast and in what manner \(f\) changes as we move in the direction of \(\mathbf{u}\).
Directional Derivative
Directional derivatives give us insight into the rate at which a function changes as we move in a specified direction. To calculate a directional derivative, we use a unit vector \(\mathbf{u}\) that signifies the direction we're interested in. In our exercise, this unit vector was \(\langle 4/\sqrt{66}, -5/\sqrt{66}, -5/\sqrt{66} \rangle\). Once the gradient, \(abla f\), is known, the directional derivative is the dot product \(abla f \cdot \mathbf{u}\). This computation measures the change of \(f\) following \(\mathbf{u}\) and indicates both the slope and sign (positive for increase, negative for decrease). In our context, the directional derivative showed the rate of descent as the function decreased in the chosen direction.
Multivariable Functions
Multivariable functions, such as \(f(x, y, z) = xy + yz^2 + xz^3\), are important in modeling scenarios with more than a single independent variable. They allow for the exploration of how changes in variables can affect outcomes differently in different dimensions. Analyzing such functions is crucial in fields like physics or economics, where the real-world phenomena depend on various inputs. For these functions, partial derivatives, gradients, and directional derivatives are tools that help understand how slight variations in one or more variables influence the behavior of the function, enabling predictions and optimizations. They are foundational concepts to tackle complex systems where each variable plays a unique role.

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Most popular questions from this chapter

Let \(f\) be the function defined by \(f(x, y)=2 x^{2}+3 y^{3}\). a. Find the equation of the tangent plane to \(f\) at the point (1,2) . b. Use the linearization to approximate the values of \(f\) at the points (1.1,2.05) and (1.3,2.2) c. Compare the approximations form part (b) to the exact values of \(f(1.1,2.05)\) and \(f(1.3,2.2) .\) Which approximation is more accurate. Explain why this should be expected.

In this section we argued that if \(f=f(x, y)\) is a function of two variables and if \(f_{x}\) and \(f_{y}\) both exist and are continuous in an open disk containing the point \(\left(x_{0}, y_{0}\right),\) then \(f\) is differentiable at \(\left(x_{0}, y_{0}\right) .\) This condition ensures that \(f\) is differentiable at \(\left(x_{0}, y_{0}\right),\) but it does not define what it means for \(f\) to be differentiable at \(\left(x_{0}, y_{0}\right) .\) In this exercise we explore the definition of differentiability of a function of two variables in more detail. Throughout, let \(g\) be the function defined by \(g(x, y)=\sqrt{|x y|}\) a. Use appropriate technology to plot the graph of \(g\) on the domain \([-1,1] \times[-1,1] .\) Explain why \(g\) is not locally linear at (0,0) b. Show that both \(g_{x}(0,0)\) and \(g_{y}(0,0)\) exist. If \(g\) is locally linear at \((0,0),\) what must be the equation of the tangent plane \(L\) to \(g\) at (0,0)\(?\) c. Recall that if a function \(f=f(x)\) of a single variable is differentiable at \(x=x_{0},\) then $$f^{\prime}\left(x_{0}\right)=\lim _{h \rightarrow 0} \frac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h}$$ exists. We saw in single variable calculus that the existence of \(f^{\prime}\left(x_{0}\right)\) means that the graph of \(f\) is locally linear at \(x=x_{0}\). In other words, the graph of \(f\) looks like its linearization \(L(x)=f\left(x_{0}\right)+\) \(f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)\) for \(x\) close to \(x_{0} .\) That is, the values of \(f(x)\) can be closely approximated by \(L(x)\) as long as \(x\) is close to \(x_{0}\). We can measure how good the approximation of \(L(x)\) is to \(f(x)\) with the error function $$E(x)=L(x)-f(x)=f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)\left(x-x_{0}\right)-f(x)$$ As \(x\) approaches \(x_{0}, E(x)\) approaches \(f\left(x_{0}\right)+f^{\prime}\left(x_{0}\right)(0)-f\left(x_{0}\right)=0\), and so \(L(x)\) provides increasingly better approximations to \(f(x)\) as \(x\) gets closer to \(x_{0} .\) Show that, even though \(g(x, y)=\sqrt{|x y|}\) is not locally linear at \((0,0),\) its error term $$ E(x, y)=L(x, y)-g(x, y) $$ at (0,0) has a limit of 0 as \((x, y)\) approaches \((0,0) .\) (Use the linearization you found in part (b).) This shows that just because an error term goes to 0 as \((x, y)\) approaches \(\left(x_{0}, y_{0}\right),\) we cannot conclude that a function is locally linear at \(\left(x_{0}, y_{0}\right)\). d. As the previous part illustrates, having the error term go to 0 does not ensure that a function of two variables is locally linear. Instead, we need a notation of a relative error. To see how this works, let us return to the single variable case for a moment and consider \(f=f(x)\) as a function of one variable. If we let \(x=x_{0}+h,\) where \(|h|\) is the distance from \(x\) to \(x_{0}\), then the relative error in approximating \(f\left(x_{0}+h\right)\) with \(L\left(x_{0}+h\right)\) is $$\frac{E\left(x_{0}+h\right)}{h}$$ Show that, for a function \(f=f(x)\) of a single variable, the limit of the relative error is 0 as \(h\) approaches 0 . e. Even though the error term for a function of two variables might have a limit of 0 at a point, our example shows that the function may not be locally linear at that point. So we use the concept of relative error to define differentiability of a function of two variables. When we consider differentiability of a function \(f=f(x, y)\) at a point \(\left(x_{0}, y_{0}\right),\) then if \(x=x_{0}+h\) and \(y=y_{0}+k,\) the distance from \((x, y)\) to \(\left(x_{0}, y_{0}\right)\) is \(\sqrt{h^{2}+k^{2}}\)

Find the differential of the function \(w=x^{3} \sin \left(y^{5} z^{1}\right)\) \(d w=\) ____________ \(d x+\) ___________ \(d y+\) \(d z\).

Suppose \(w=\frac{x}{y}+\frac{y}{z},\) where $$x=e^{5 t}, y=2+\sin (3 t), \text { and } z=2+\cos (6 t)$$ A) Use the chain rule to find \(\frac{d w}{d t}\) as a function of \(x, y, z,\) and \(t .\) Do not rewrite \(x, y,\) and \(z\) in terms of \(t,\) and do not rewrite \(e^{5 t}\) as \(x\). \(\frac{d w}{d t}=\) ____________. B) Use part A to evaluate \(\frac{d w}{\|}\) when \(t=0\).

If \(f(x, y)=3 x^{2}-1 y^{2},\) find the value of the directional derivative at the point (-1,-4) in the direction given by the angle \(\theta=\frac{2 \pi}{6}\).

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