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Find the derivative of the function at \(P_{0}\) in the direction of \(\mathbf{u}.\) \(\begin{array}{l}{h(x, y, z)=\cos x y+e^{y z}+\ln z x, \quad P_{0}(1,0,1 / 2)} \\\ {\mathbf{u}=\mathbf{i}+2 \mathbf{j}+2 \mathbf{k}}\end{array}\)

Short Answer

Expert verified
The directional derivative is \( \frac{5}{3} \).

Step by step solution

01

Directional Derivative Formula

To find the directional derivative of a function \( h(x, y, z) \) in the direction of a vector \( \mathbf{u} \), use the formula: \( D_{\mathbf{u}}h = abla h \cdot \mathbf{u} \), where \( abla h \) is the gradient of \( h \) and \( \mathbf{u} \) is a unit vector.
02

Find the Gradient

Compute the partial derivatives: \( \frac{\partial h}{\partial x} = -y\sin xy + \frac{1}{zx}, \frac{\partial h}{\partial y} = -x\sin xy + ze^{yz}, \frac{\partial h}{\partial z} = y e^{yz} + \frac{x}{z} \), then evaluate at \( P_0(1,0,\frac{1}{2}) \). This gives \( abla h(1,0,\frac{1}{2}) = \left(0, \frac{1}{2}, 2 \right) \).
03

Normalize the Vector \( \mathbf{u} \)

Normalize \( \mathbf{u} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \). The magnitude of \( \mathbf{u} \) is \( \| \mathbf{u} \| = \sqrt{1^2 + 2^2 + 2^2} = 3 \). Hence, the unit vector \( \mathbf{u_{unit}} = \left(\frac{1}{3}, \frac{2}{3}, \frac{2}{3} \right) \).
04

Compute the Dot Product

Calculate the dot product \( abla h \cdot \mathbf{u_{unit}} = \left(0, \frac{1}{2}, 2 \right) \cdot \left(\frac{1}{3}, \frac{2}{3}, \frac{2}{3} \right) = 0 \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{2}{3} + 2 \cdot \frac{2}{3} \).
05

Solve the Dot Product

Calculate each term: \( 0 + \frac{1}{3} + \frac{4}{3} = \frac{5}{3} \). Therefore, \( D_{\mathbf{u}}h(1,0, \frac{1}{2}) = \frac{5}{3} \).

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Key Concepts

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

Directional Derivative
The directional derivative is a way to understand how a function changes as you move in a specific direction from a certain point. Imagine you are standing on a hill and want to know how steep it is in the direction you are heading. The directional derivative tells you this steepness for any path you choose.

To calculate the directional derivative of a function, you need two things:
  • The gradient of the function at the point of interest.
  • A unit vector that represents the direction you want to move.
Once you have both, you take the dot product of the gradient vector and the unit vector. This result gives you the directional derivative.
Gradient Vector
The gradient vector is like a compass pointing in the direction of the steepest ascent of a function. If you think of a landscape, the gradient vector tells you the direction to walk in to reach the highest point the fastest.

This vector is composed of partial derivatives, which are the rates of change of the function with respect to each variable. For a function expressed as \( h(x, y, z) \), the gradient is calculated as:
  • \( abla h = \left( \frac{\partial h}{\partial x}, \frac{\partial h}{\partial y}, \frac{\partial h}{\partial z} \right) \)
Evaluating the gradient at a specific point gives you the slope information precisely at that location.
Partial Derivatives
Partial derivatives are the building blocks of gradient vectors. They describe how a function changes as each individual variable changes, while keeping the others constant. Think of them as slices of the function that reveal the characteristics of its surface.

To compute a partial derivative, differentiate the function only with respect to one variable at a time. For example, for \( h(x, y, z) \):
  • \( \frac{\partial h}{\partial x} \): How \( h \) changes as \( x \) changes.
  • \( \frac{\partial h}{\partial y} \): How \( h \) changes as \( y \) changes.
  • \( \frac{\partial h}{\partial z} \): How \( h \) changes as \( z \) changes.
These derivatives are essential for creating the gradient vector.
Vector Normalization
Vector normalization is the process of converting a vector into a unit vector, meaning it has a length or magnitude of one but points in the same direction. This is crucial because, in calculations like the directional derivative, using a unit vector ensures that the resulting value only represents rate of change and not the influence of the vector's magnitude.

To normalize a vector, you divide each component by the vector's magnitude. For instance, if \( \mathbf{u} = \mathbf{i} + 2\mathbf{j} + 2\mathbf{k} \, \) its magnitude \( \| \mathbf{u} \| = \sqrt{1^2 + 2^2 + 2^2} = 3 \). So, the unit vector \( \mathbf{u_{unit}} \) is:
  • \( \mathbf{u_{unit}} = \left( \frac{1}{3}, \frac{2}{3}, \frac{2}{3} \right) \)
This ensures computations such as the dot product for directional derivatives are correct.

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Most popular questions from this chapter

Find the value of \(\partial x / \partial z\) at the point \((1,-1,-3)\) if the equation $$x z+y \ln x-x^{2}+4=0$$ defines \(x\) as a function of the two independent variables \(y\) and \(z\) and the partial derivative exists.

Minimize the function \(f(x, y, z)=x^{2}+y^{2}+z^{2}\) subject to the constraints \(x+2 y+3 z=6\) and \(x+3 y+9 z=9\) .

Use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema: a. Form the function \(h=f-\lambda_{1} g_{1}-\lambda_{2} g_{2},\) where \(f\) is the function to optimize subject to the constraints \(g_{1}=0\) and \(g_{2}=0\) . b. Determine all the first partial derivatives of \(h\) , including the partials with respect to \(\lambda_{1}\) and \(\lambda_{2},\) and set them equal to \(0 .\) c. Solve the system of equations found in part (b) for all the unknowns, including \(\lambda_{1}\) and \(\lambda_{2}\) . d. Evaluate \(f\) at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the exercise. Minimize \(f(x, y, z)=x^{2}+y^{2}+z^{2}\) subject to the constraints \(x^{2}-x y+y^{2}-z^{2}-1=0\) and \(x^{2}+y^{2}-1=0.\)

Hottest point on a space probe A space probe in the shape of the ellipsoid $$4 x^{2}+y^{2}+4 z^{2}=16$$ enters Earth's atmosphere and its surface begins to heat. After 1 hour, the temperature at the point \((x, y, z)\) on the probe's surface is $$T(x, y, z)=8 x^{2}+4 y z-16 z+600.$$ Find the hottest point on the probe's surface.

If a function \(f(x, y)\) has continuous second partial derivatives throughout an open region \(R,\) must the first-order partial derivatives of \(f\) be continuous on \(R ?\) Give reasons for your answer.

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