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Suppose \(f\) is differentiable at \((3,4), \nabla f(3,4)=\langle-\sqrt{3}, 1\rangle\) and \(\mathbf{u}=\left\langle\frac{\sqrt{3}}{2},-\frac{1}{2}\right\rangle .\) Compute \(D_{\mathbf{u}} f(3,4)\).

Short Answer

Expert verified
Based on the given gradient \(\nabla f(3,4)=\langle-\sqrt{3},1\rangle\) and the directional vector \(\mathbf{u}=\left\langle\frac{\sqrt{3}}{2},-\frac{1}{2}\right\rangle\), we calculated the directional derivative \(D_{\mathbf{u}} f(3,4)\) using the formula \(D_{\mathbf{u}} f(3,4) = \nabla f(3,4) \cdot \mathbf{u}\), which simplifies to \(D_{\mathbf{u}} f(3,4) = -2\).

Step by step solution

01

Write down the formula for the directional derivative

The formula for the directional derivative is \(D_{\mathbf{u}} f(3,4) = \nabla f(3,4) \cdot \mathbf{u}\). We need to calculate the dot product between the given gradient and the given directional vector.
02

Substitute the given values into the formula

We have \(\nabla f(3,4)=\langle-\sqrt{3},1\rangle\) and \(\mathbf{u}=\left\langle\frac{\sqrt{3}}{2},-\frac{1}{2}\right\rangle\). Substitute these values into the formula: \(D_{\mathbf{u}} f(3,4) = \langle-\sqrt{3},1\rangle \cdot \left\langle\frac{\sqrt{3}}{2},-\frac{1}{2}\right\rangle\).
03

Calculate the dot product

Calculate the dot product between \(\langle-\sqrt{3},1\rangle\) and \(\left\langle\frac{\sqrt{3}}{2},-\frac{1}{2}\right\rangle\): \(D_{\mathbf{u}} f(3,4) = (-\sqrt{3})\left(\frac{\sqrt{3}}{2}\right) + (1)\left(-\frac{1}{2}\right) = -\frac{3}{2}-\frac{1}{2}\).
04

Simplify the expression

Simplify the expression for the directional derivative: \(D_{\mathbf{u}} f(3,4) = -\frac{3+1}{2} = -\frac{4}{2} = -2\). The directional derivative \(D_{\mathbf{u}} f(3,4)\) is equal to -2.

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Key Concepts

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

Gradient Vector
The gradient vector is a crucial concept in vector calculus and plays a pivotal role in understanding directional derivatives. It is denoted by \( abla f \) and helps to determine the direction of the steepest ascent of a function. For a function \( f \) with variables \( x \) and \( y \), the gradient is represented as:
  • \( abla f(x, y) = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle \)
In our exercise, the gradient vector at the point (3,4) is given as \( abla f(3,4) = \langle -\sqrt{3}, 1 \rangle \). This means if we move in the direction of this vector from the point (3,4), the function \( f \) will increase at the maximum possible rate. This vector essentially summarizes information about the slope of \( f \) in every possible direction.
Dot Product
The dot product, also known as the scalar product, takes two equal-length sequences of numbers and returns a single number. It is foundational to calculating directional derivatives. For vectors \( \mathbf{a} = \langle a_1, a_2 \rangle \) and \( \mathbf{b} = \langle b_1, b_2 \rangle \), the dot product is calculated as:
  • \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 \)
In the context of our problem, we use the dot product to find \( D_{\mathbf{u}} f(3,4) \), where we calculate it between \( abla f(3,4) = \langle -\sqrt{3}, 1 \rangle \) and the direction vector \( \mathbf{u} = \left\langle \frac{\sqrt{3}}{2}, -\frac{1}{2} \right\rangle \). This step is critical as it quantifies the change in the function's value as we move in the direction of \( \mathbf{u} \). Calculating \(-\sqrt{3} \cdot \frac{\sqrt{3}}{2} + 1 \cdot -\frac{1}{2}\) gives us \(-\frac{4}{2} = -2\).
Vector Calculus
Vector calculus is an area of mathematics focusing on the differentiation and integration of vector fields. It extends the concepts of calculus to describe physical phenomena more precisely. Utilizing vector calculus principles, we can explore variations in scalar fields such as temperature or pressure.In the scenario of directional derivatives, vector calculus allows us to measure how a function changes as we move along any direction. The directional derivative \( D_{\mathbf{u}} f \) is a powerful tool here. It uses the capabilities of vectors and calculus to give insight into the behavior of functions in multiple dimensions.The given exercise effectively ties together these concepts to find how rapidly a function changes in a specific direction. This exploration helps develop a deeper understanding of multi-variable functions and prepares students for complex real-world problems.

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