By about how much will

f left parenthesis x comma y comma z right parenthesisf(x,y,z)equals=ln StartRoot x squared plus y squared plus z squared EndRootlnx2+y2+z2

change if the point

Upper P left parenthesis x comma y comma z right parenthesisP(x,y,z)

moves from

Upper P 0 left parenthesis 6 comma 5 comma 15 right parenthesisP0(6,5,15)

a distance of

dsdsequals=0.10.1

unit in the direction of

3 Bold i plus 6 Bold j minus 2 Bold k3**i**+6**j**−2**k**​?  round final answer to 4 decimal places

To solve this problem, we need to compute the differential change in the function \( f(x, y, z) = \ln \sqrt{x^2 + y^2 + z^2} \) when the point \( P(x, y, z) \) moves from \( P_0(6, 5, 15) \) by a small distance \( ds = 0.1 \) in the direction of \( \mathbf{v} = 3\mathbf{i} + 6\mathbf{j} - 2\mathbf{k} \).

### Step 1: Gradient of \( f(x, y, z) \)
The gradient of \( f(x, y, z) \) gives us the direction and rate of change of \( f \) at any point:

\[
\nabla f(x, y, z) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)
\]

First, let's rewrite the function \( f(x, y, z) \):

\[
f(x, y, z) = \ln \sqrt{x^2 + y^2 + z^2} = \frac{1}{2} \ln (x^2 + y^2 + z^2)
\]

The partial derivatives of \( f(x, y, z) \) are:

\[
\frac{\partial f}{\partial x} = \frac{x}{x^2 + y^2 + z^2}, \quad \frac{\partial f}{\partial y} = \frac{y}{x^2 + y^2 + z^2}, \quad \frac{\partial f}{\partial z} = \frac{z}{x^2 + y^2 + z^2}
\]

Thus, the gradient is:

\[
\nabla f(x, y, z) = \left( \frac{x}{x^2 + y^2 + z^2}, \frac{y}{x^2 + y^2 + z^2}, \frac{z}{x^2 + y^2 + z^2} \right)
\]

### Step 2: Evaluate the gradient at \( P_0(6, 5, 15) \)
At the point \( P_0(6, 5, 15) \), we need to evaluate \( \nabla f \):

\[
x_0 = 6, \quad y_0 = 5, \quad z_0 = 15
\]

\[
x_0^2 + y_0^2 + z_0^2 = 6^2 + 5^2 + 15^2 = 36 + 25 + 225 = 286
\]

Now, the components of \( \nabla f \) at \( P_0 \) are:

\[
\frac{\partial f}{\partial x}\bigg|_{P_0} = \frac{6}{286}, \quad \frac{\partial f}{\partial y}\bigg|_{P_0} = \frac{5}{286}, \quad \frac{\partial f}{\partial z}\bigg|_{P_0} = \frac{15}{286}
\]

Thus, the gradient at \( P_0 \) is:

\[
\nabla f(6, 5, 15) = \left( \frac{6}{286}, \frac{5}{286}, \frac{15}{286} \right)
\]

### Step 3: Direction of movement
The direction of movement is given by the vector \( \mathbf{v} = 3\mathbf{i} + 6\mathbf{j} - 2\mathbf{k} \). We first normalize this direction vector:

\[
|\mathbf{v}| = \sqrt{3^2 + 6^2 + (-2)^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7
\]

So, the unit direction vector is:

\[
\hat{\mathbf{v}} = \left( \frac{3}{7}, \frac{6}{7}, \frac{-2}{7} \right)
\]

### Step 4: Dot product of \( \nabla f \) and the direction vector
The rate of change of \( f \) in the direction of \( \mathbf{v} \) is given by the directional derivative, which is the dot product of \( \nabla f \) and the unit direction vector \( \hat{\mathbf{v}} \):

\[
\nabla f \cdot \hat{\mathbf{v}} = \frac{6}{286} \cdot \frac{3}{7} + \frac{5}{286} \cdot \frac{6}{7} + \frac{15}{286} \cdot \frac{-2}{7}
\]

Let's compute this:

\[
\nabla f \cdot \hat{\mathbf{v}} = \frac{18}{2002} + \frac{30}{2002} - \frac{30}{2002} = \frac{18}{2002}
\]

\[
\nabla f \cdot \hat{\mathbf{v}} = \frac{18}{2002} \approx 0.00899
\]

### Step 5: Compute the change in \( f \)
The change in \( f \) when the point moves by \( ds = 0.1 \) is approximately:

\[
df \approx \nabla f \cdot \hat{\mathbf{v}} \cdot ds = 0.00899 \times 0.1 = 0.000899
\]

### Final Answer
The approximate change in \( f \) is \( \boxed{0.0009} \) (rounded to four decimal places).

---

Would you like more details on any step? Here are five related questions:

1. How is the directional derivative conceptually different from a regular derivative?
2. How do we compute gradients for functions with more than three variables?
3. What are other methods to estimate the change in a multivariable function?
4. Can directional derivatives be negative, and what does that signify?
5. How would this problem change if the direction vector was normalized differently?

**Tip:** The gradient vector always points in the direction of the steepest ascent of the function.

By about how much will f(x, y, z) = ln(sqrt(x^2 + y^2 + z^2)) change if the point P(x, y, z) moves from P_0(6, 5, 15) a distance of ds = 0.1 unit in the direction of 3i + 6j - 2k? Round final answer to 4 decimal places.

Explore the directional derivative of f(x, y, z) = ln(sqrt(x^2 + y^2 + z^2)) when moving from point P_0(6, 5, 15) by a small distance of 0.1 unit in the direction of 3i + 6j - 2k. Learn about gradients and directional derivatives in multivariable calculus with step-by-step solutions.

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