Problem 12 Find the derivative of the funct... [FREE SOLUTION]

Chapter 14: Problem 12

Find the derivative of the function at $P_{0}$ in the direction of $\mathbf{u}.$ $$f(x, y)=2 x^{2}+y^{2}, \quad P_{0}(-1,1), \quad \mathbf{u}=3 \mathbf{i}-4 \mathbf{j}.$$

Short Answer

Expert verified

The derivative of the function at $P_0$ in the direction of $\mathbf{u}$ is $-4$.

Step by step solution

Compute the Gradient of the Function

For a function $f(x, y) = 2x^2 + y^2$, the gradient $abla f$ is obtained by partially differentiating with respect to $x$ and $y$. Compute these derivatives:$\frac{\partial f}{\partial x} = 4x$ and $\frac{\partial f}{\partial y} = 2y$.Thus, the gradient is $abla f = \langle 4x, 2y \rangle$.

Evaluate the Gradient at the Point $P_0$

Substitute the coordinates of $P_0(-1,1)$ into the gradient:$abla f(-1, 1) = \langle 4(-1), 2(1) \rangle = \langle -4, 2 \rangle$.

Normalize the Direction Vector $\mathbf{u}$

The direction vector given is $\mathbf{u} = 3\mathbf{i} - 4\mathbf{j}$. First, calculate its magnitude:$\|\mathbf{u}\| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.Normalize $\mathbf{u}$ by dividing by its magnitude:$\hat{\mathbf{u}} = \left\langle \frac{3}{5}, \frac{-4}{5} \right\rangle$.

Compute the Directional Derivative

The directional derivative of $f$ at $P_0(-1,1)$ in the direction of $\hat{\mathbf{u}}$ is given by the dot product $abla f(P_0) \cdot \hat{\mathbf{u}}$.$abla f(-1, 1) = \langle -4, 2 \rangle$ and $\hat{\mathbf{u}} = \left\langle \frac{3}{5}, \frac{-4}{5} \right\rangle$.Compute the dot product:$-4 \cdot \frac{3}{5} + 2 \cdot \frac{-4}{5} = -\frac{12}{5} + \left(-\frac{8}{5}\right) = -\frac{20}{5} = -4$.

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Key Concepts

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

Partial Derivatives

When dealing with functions of several variables, a partial derivative is a way to find how the function changes, not with all its variables, but with respect to one variable at a time. This means we hold all other variables constant while finding the derivative.Partial derivatives help us analyze changes in multi-variable functions:

To find a partial derivative with respect to variable x, treat y as a constant and differentiate.
Similarly, for variable y, treat x as a constant.

For example, with the function given in the exercise, the partial derivative with respect to x gives the rate of change of the function when x changes and y stays the same, resulting in $\frac{\partial f}{\partial x} = 4x$. Meanwhile, $\frac{\partial f}{\partial y} = 2y$ is used when only y changes.

Gradient Vector

In calculus, the gradient vector represents the direction and rate of the steepest increase of a function. For a function of two variables, the gradient is a vector which:

Combines the partial derivatives along the x and y directions.
Is notated as $abla f$ which equals $\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \rangle$.

In our case, with the function $f(x, y) = 2x^2 + y^2$, the gradient vector is $\langle 4x, 2y \rangle$. When you insert the point of interest $P_0(-1,1)$, you evaluate the gradient to $\langle -4, 2 \rangle$. This vector tells us where, at that point, the function changes the most and at what rate.

Vector Normalization

Vector normalization transforms a vector to have a magnitude of one but still point in the same direction. This is crucial for understanding directional derivatives because:

The unit vector gives the precise direction of interest.
It maintains the direction while scaling down the size to unity.

Given the direction vector $\mathbf{u} = 3\mathbf{i} - 4\mathbf{j}$, the magnitude is calculated as $\sqrt{3^2 + (-4)^2} = 5$. We obtain the normalized vector by dividing each component by 5, resulting in $\hat{\mathbf{u}} = \langle \frac{3}{5}, \frac{-4}{5} \rangle$. Using this, we can accurately compute the directional derivative aligned perfectly with the vector's intended direction while ensuring a standardized magnitude.

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91影视

Find the derivative of the function at \(P_{0}\) in the direction of \(\mathbf{u}.\) $$f(x, y)=2 x^{2}+y^{2}, \quad P_{0}(-1,1), \quad \mathbf{u}=3 \mathbf{i}-4 \mathbf{j}.$$

Short Answer

Step by step solution

Compute the Gradient of the Function

Evaluate the Gradient at the Point \(P_0\)

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

Compute the Directional Derivative

Key Concepts

Partial Derivatives

Gradient Vector

Vector Normalization

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