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\(4-6\) Find the directional derivative of \(f\) at the given point in the direction indicated by the angle \(\theta .\) $$f(x, y)=x^{2} y^{3}-y^{4}, \quad(2,1), \quad \theta=\pi / 4$$

Short Answer

Expert verified
The directional derivative is \(6\sqrt{2}\).

Step by step solution

01

Find the Gradient of the Function

The gradient of a function \(f(x, y)\) is the vector of its partial derivatives. For \(f(x, y) = x^2y^3 - y^4\), calculate the partial derivatives:- Partial derivative with respect to \(x\):\[ f_x(x, y) = \frac{\partial}{\partial x}(x^2y^3 - y^4) = 2xy^3 \]- Partial derivative with respect to \(y\):\[ f_y(x, y) = \frac{\partial}{\partial y}(x^2y^3 - y^4) = 3x^2y^2 - 4y^3 \]Thus, the gradient is:\[ abla f(x, y) = \langle 2xy^3, 3x^2y^2 - 4y^3 \rangle \]
02

Evaluate the Gradient at the Given Point

Substitute the point \((2, 1)\) into the gradient. - Evaluate \(f_x(2, 1): 2 \times 2 \times 1^3 = 4\)- Evaluate \(f_y(2, 1): 3 \times 2^2 \times 1^2 - 4 \times 1^3 = 12 - 4 = 8\)The gradient at \((2, 1)\) is:\[ abla f(2, 1) = \langle 4, 8 \rangle \]
03

Determine the Direction Vector Using Angle \(\theta\)

The direction is given by the angle \(\theta = \frac{\pi}{4}\). The direction vector based on this angle is \(\mathbf{u} = \langle \cos(\pi/4), \sin(\pi/4) \rangle\).- \(\cos(\pi/4) = \sin(\pi/4) = \frac{\sqrt{2}}{2}\), so\[ \mathbf{u} = \left\langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle \]
04

Compute the Dot Product for the Directional Derivative

The directional derivative \(D_\mathbf{u}f(2,1)\) in the direction \(\mathbf{u}\) is given by the dot product of the gradient and the direction vector: \[ D_\mathbf{u}f(2, 1) = abla f(2, 1) \cdot \mathbf{u} \]Calculate the dot product:\[ D_\mathbf{u}f(2, 1) = \langle 4, 8 \rangle \cdot \left\langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle \]\[ = 4 \cdot \frac{\sqrt{2}}{2} + 8 \cdot \frac{\sqrt{2}}{2} \]\[ = 2\sqrt{2} + 4\sqrt{2} = 6\sqrt{2} \]
05

Conclusion

The directional derivative of the function at the point \((2, 1)\) in the direction given by the angle \(\theta = \frac{\pi}{4}\) is \(6\sqrt{2}\).

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Key Concepts

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

Gradient
The concept of the Gradient is pivotal when exploring functions of multiple variables in calculus. Essentially, the gradient symbolizes the direction of the steepest ascent of a function. It is represented as a vector consisting of all the partial derivatives with respect to each variable.

To illustrate, consider a function of two variables, such as \( f(x, y) = x^2y^3 - y^4 \). The gradient, denoted as \( abla f(x, y) \), is formed by:
  • The partial derivative with respect to \( x \), \( f_x(x, y) = 2xy^3\).
  • The partial derivative with respect to \( y \), \( f_y(x, y) = 3x^2y^2 - 4y^3 \).
Thus, the gradient vector here is \( abla f(x, y) = \langle 2xy^3, 3x^2y^2 - 4y^3 \rangle \).

This vector not only directs the rate of change of the function but also gives the magnitude of change. It's particularly helpful for optimization problems, as it shows how to adjust the input variables for maximum benefit.
Partial Derivative
When working on functions with multiple variables, Partial Derivatives help us understand how the function changes with respect to one variable, while keeping others constant. It's akin to holding other measurements static while seeing how one alone influences the outcome.

For the function \( f(x, y) = x^2y^3 - y^4 \), let's determine its partial derivatives:
  • Regarding \( x \), the partial derivative is \( f_x(x, y) = 2xy^3 \). Notice how \( y \) is treated as a constant here.
  • For \( y \), the partial derivative is \( f_y(x, y) = 3x^2y^2 - 4y^3 \), where \( x \) is held constant.
These derivatives are crucial, especially when constructing gradient vectors. They tell us how sensitive the function is with changes in each respective variable. In calculus and multidimensional analysis, this becomes essential for identifying critical points and analyzing behavior within functions.
Dot Product
Dot Product is an essential operation in vector mathematics, connecting vectors to quantities. In the context of directional derivatives, it helps establish the rate of change of the function in a specified direction.

To find the dot product, we multiply corresponding components of two vectors and sum the results. Let's apply this principle:
  • Considering the gradient vector \( abla f(2, 1) = \langle 4, 8 \rangle \).
  • And, the direction vector derived from angle \( \theta \), \( \mathbf{u} = \left\langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle \).
The dot product, thus, is calculated as:
\[ D_\mathbf{u}f(2, 1) = \langle 4, 8 \rangle \cdot \left\langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle = 2\sqrt{2} + 4\sqrt{2} = 6\sqrt{2} \]

This result signifies the directional derivative, indicating the function's rate of change in that particular direction. It's a fundamental technique used broadly in physics and engineering fields for analyzing vector quantities.
Angle in Radians
Angles in radians provide a natural way to express angles in calculus due to their relation to the radius of a circle. When dealing with directional derivatives, specifying the direction often involves angles like \( \theta = \frac{\pi}{4} \).

Understanding radians is crucial:
  • One radian corresponds to the angle formed when the arc length is equal to the radius of a circle.
  • In situations involving trigonometry, such as calculating direction vectors, using radians simplifies the computation.
For instance, to find a direction vector using \( \theta = \frac{\pi}{4} \), use:
  • \( \cos(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \)
  • \( \sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2} \)
Thus, the vector is \( \left\langle \frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2} \right\rangle \). Using radians keeps the math rigorous and precise, allowing for a seamless connection between angle measures and distance units.

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Most popular questions from this chapter

The kinetic energy of a body with mass \(m\) and velocity \(v\) is \(K=\frac{1}{2} m v^{2} .\) Show that $$\frac{\partial K}{\partial m} \frac{\partial^{2} K}{\partial v^{2}}=K$$

Find the absolute maximum and minimum values of \(f\) on the set \(D .\) \(f(x, y)=1+4 x-5 y, \quad D\) is the closed triangular region with vertices \((0,0),(2,0),\) and \((0,3)\)

Suppose that a scientist has reason to believe that two quan- tities \(x\) and \(y\) are related linearly, that is, \(y=m x+b,\) at least approximately, for some values of \(m\) and \(b\) . The scientist performs an experiment and collects data in the form of points \(\left(X_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)\) and then plots these points. The points don't lie exactly on a straight line, so the scientist wants $$ m \text { and } b \text { so that the line } y=m x+b$$ points as well as possible. (See the figure.) Let \(d_{i}=y_{i}-\left(m x_{i}+b\right)\) be the vertical deviation of the point \(\left(x_{i}, y_{i}\right)\) from the line. The method of least squares determines \(m\) and \(b\) so as to minimize \(\Sigma_{1-1}^{n} d_{i}^{2}\) , the sum of the squares of these deviations. Show that, according to this method, the line of best fit is obtained when $$\begin{array}{c}{m \sum_{i=1}^{n} x_{i}+b n=\sum_{i=1}^{n} y_{i}} \\ {m \sum_{i=1}^{n} x_{i}^{2}+b \sum_{i=1}^{n} x_{i}=\sum_{i=1}^{n} x_{i} y_{i}}\end{array}$$ Thus the line is found by solving these two equations in the two unknowns \(m\) and \(b\) . See Section 1.2 for a further discus- sion and applications of the method of least squares.)

Find the absolute maximum and minimum values of \(f\) on the set \(D .\) $$f(x, y)=2 x^{3}+y^{4}, \quad D=\\{(x, y) | x^{2}+y^{2} \leqslant 1\\}$$

Find the maximum and minimum values of \(f\) subject to the given constraints. Use a computer algebra system to solve the system of equations that arises in using Lagrange multipliers. (If your CAS finds only one solution, you may need to use additional commands.) \(f(x, y, z)=x+y+z ; \quad x^{2}-y^{2}=z, x^{2}+z^{2}=4\)

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