The gradient is an important concept in vector calculus. It helps us understand how a scalar function changes with respect to different variables. In three-dimensional space, the gradient is calculated by finding the partial derivatives of a function with respect to x, y, and z.
In this question, we need to prove a mathematical identity involving the natural logarithm of the magnitude of a position vector.
Gradient of ln r: Question
Question: Let ( \mathbf{r}=\langle x,y,z\rangle ). Prove that:
[
\nabla \ln |\mathbf{r}| =
\frac{\mathbf{r}}{|\mathbf{r}|^2}
]
Answer and Explanation
Answer: The given identity can be proved using the definition of the gradient and the partial derivatives of a function.
Step 1: Find the Magnitude of the Vector
Given:
[
\mathbf{r}=\langle x,y,z\rangle
]
The magnitude of this vector is:
[
|\mathbf{r}|=\sqrt{x^2+y^2+z^2}
]
Taking the natural logarithm on both sides:
[
\ln|\mathbf{r}| =
\ln\sqrt{x^2+y^2+z^2}
]
Using the logarithmic property:
[
f(x,y,z)=\frac{1}{2}\ln(x^2+y^2+z^2)
]
Step 2: Apply the Gradient Formula
The gradient of a scalar function in three dimensions is defined as:
[
\nabla f =
\left\langle
\frac{\partial f}{\partial x},
\frac{\partial f}{\partial y},
\frac{\partial f}{\partial z}
\right\rangle
]
Now, we calculate the partial derivatives separately.
Step 3: Calculate the Partial Derivatives
Partial derivative with respect to x:
\frac{1}{2}\cdot
\frac{2x}{x^2+y^2+z^2}
]
[
=\frac{x}{x^2+y^2+z^2}
]
Partial derivative with respect to y:
\frac{1}{2}\cdot
\frac{2y}{x^2+y^2+z^2}
]
[
=\frac{y}{x^2+y^2+z^2}
]
Partial derivative with respect to z:
\frac{1}{2}\cdot
\frac{2z}{x^2+y^2+z^2}
]
[
=\frac{z}{x^2+y^2+z^2}
]
Step 4: Combine the Results
Now, putting all three partial derivatives together:
[
\nabla f =
\left\langle
\frac{x}{x^2+y^2+z^2},
\frac{y}{x^2+y^2+z^2},
\frac{z}{x^2+y^2+z^2}
\right\rangle
]
Taking the common denominator:
[
\nabla f =
\frac{\langle x,y,z\rangle}
{x^2+y^2+z^2}
]
We know that:
[
\mathbf{r}=\langle x,y,z\rangle
]
And:
[
|\mathbf{r}|^2=x^2+y^2+z^2
]
Therefore:
[
\boxed{
\nabla\ln|\mathbf{r}|=
\frac{\mathbf{r}}{|\mathbf{r}|^2}
}
]
Final Answer
Hence, we have proved that:
[
\boxed{
\nabla\ln|\mathbf{r}|=
\frac{\mathbf{r}}{|\mathbf{r}|^2}
}
]
Therefore, the given identity is proved for all nonzero position vectors ((\mathbf{r}\neq 0)).
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