Chapter 8: Problem 12
The directional derivative of \(\phi(x, y, z)=x^{2} y z+4 x^{2}\) at the point \((1,-2,-1)\) in the direction \(F Q\) where \(P=(1,2,-1)\) and \(Q=(-1,2,3)\) is
Short Answer
Expert verified
The directional derivative is \(-\frac{16\sqrt{5}}{5}\).
Step by step solution
01
Find the Gradient of the Function
First, we need to find the gradient of the function \( \phi(x, y, z) = x^2 y z + 4x^2 \). The gradient, \( abla \phi \), is the vector of partial derivatives: \( abla \phi(x, y, z) = \left( \frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z} \right) \). Calculate each partial derivative: \( \frac{\partial \phi}{\partial x} = 2xy z + 8x \)\( \frac{\partial \phi}{\partial y} = x^2 z \)\( \frac{\partial \phi}{\partial z} = x^2 y \). So, \( abla \phi = (2xy z + 8x, x^2 z, x^2 y) \).
02
Evaluate the Gradient at the Point
Next, evaluate the gradient at the given point \( (1, -2, -1) \):\( abla \phi(1, -2, -1) = (2(1)(-2)(-1) + 8(1), (1)^2(-1), (1)^2(-2)) \). Simplify:\( abla \phi(1, -2, -1) = (4 + 8, -1, -2) = (12, -1, -2) \).
03
Find the Direction Vector and its Unit Vector
The direction vector from \( P(1, 2, -1) \) to \( Q(-1, 2, 3) \) is \( \overrightarrow{PQ} = (-1-1, 2-2, 3-(-1)) = (-2, 0, 4) \). Find its magnitude: \( |\overrightarrow{PQ}| = \sqrt{(-2)^2 + 0^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \).The unit vector \( \hat{u} \) in the direction of \( \overrightarrow{PQ} \) is:\( \hat{u} = \frac{1}{2\sqrt{5}}(-2, 0, 4) = \left(-\frac{2}{2\sqrt{5}}, 0, \frac{4}{2\sqrt{5}}\right) = \left(-\frac{1}{\sqrt{5}}, 0, \frac{2}{\sqrt{5}}\right) \).
04
Calculate the Directional Derivative
The directional derivative of \( \phi \) at \( (1, -2, -1) \) in the direction of \( \hat{u} \) is given by the dot product of \( abla \phi(1, -2, -1) \) and \( \hat{u} \):\( D_{\hat{u}}\phi = abla \phi(1, -2, -1) \cdot \hat{u} = (12, -1, -2) \cdot \left(-\frac{1}{\sqrt{5}}, 0, \frac{2}{\sqrt{5}}\right) \).Calculate the dot product: \( D_{\hat{u}}\phi = 12\left(-\frac{1}{\sqrt{5}}\right) + (-1)(0) + (-2)\left(\frac{2}{\sqrt{5}}\right) \).Simplify:\( D_{\hat{u}}\phi = -\frac{12}{\sqrt{5}} - \frac{4}{\sqrt{5}} = -\frac{16}{\sqrt{5}} \).
05
Simplify the Result
The directional derivative can also be presented in a more standard form by rationalizing the denominator:\( D_{\hat{u}}\phi = -\frac{16}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} = -\frac{16\sqrt{5}}{5} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Gradient of a Function
The gradient of a function is an essential concept in multivariable calculus. It is like a compass showing you the direction of the steepest ascent of a function. For a function of several variables, the gradient is represented as a vector. This vector comprises the partial derivatives of the function.
The gradient gives us important insight:
In our example, these partial derivatives provide:
The gradient gives us important insight:
- Direction of greatest increase of the function
- Magnitude reflects the rate of increase
In our example, these partial derivatives provide:
- \( \frac{\partial \phi}{\partial x} = 2xy z + 8x \)
- \( \frac{\partial \phi}{\partial y} = x^2 z \)
- \( \frac{\partial \phi}{\partial z} = x^2 y \)
Partial Derivatives
Partial derivatives are a fundamental component in understanding changes in multivariable functions. A partial derivative of a function at a point gives the rate at which the function changes as one of the input variables changes, with all other variables held constant.
This is pivotal because, unlike single-variable functions that vary along one axis, multivariable functions can change along multiple axes, giving rise to a partial derivative for each variable.
This is pivotal because, unlike single-variable functions that vary along one axis, multivariable functions can change along multiple axes, giving rise to a partial derivative for each variable.
- Consider \( \phi(x, y, z) = x^2 y z + 4x^2 \). Here, to find the partial derivative with respect to \( x \), treat \( y \) and \( z \) as constants: \( \frac{\partial \phi}{\partial x} = 2xy z + 8x \).
- For \( y \), treat \( x \) and \( z \) as constants: \( \frac{\partial \phi}{\partial y} = x^2 z \).
- And for \( z \), keep \( x \) and \( y \) constant: \( \frac{\partial \phi}{\partial z} = x^2 y \).
Unit Vector
In the context of directional derivatives, a unit vector plays a crucial role. A vector becomes a unit vector when it has a magnitude of exactly one. It is used to indicate direction without scaling effects of different magnitudes. A unit vector helps ensure we're only considering the direction, not the length of the vector.
To convert a vector to a unit vector:
To convert a vector to a unit vector:
- Find its magnitude. For example, the vector \( \overrightarrow{PQ} = (-2, 0, 4) \) has a magnitude of \( \sqrt{(-2)^2 + 0^2 + 4^2} = \sqrt{20} \).
- Divide each component of the vector by this magnitude, resulting in \( \hat{u} = \left(-\frac{1}{\sqrt{5}}, 0, \frac{2}{\sqrt{5}}\right) \).