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Math  /  Calculus

QuestionFind the directional derivative of the function f(x,y,z)=xy2z3 f(x, y, z) = xy^2z^3 at the point P(2,1,1) P(2,1,1) in the direction of the point Q(0,−3,5) Q(0,-3,5) .

Studdy Solution

STEP 1

What is this asking? We need to find how much the function f(x,y,z)f(x, y, z) is changing at a specific point PP if we move in the direction towards another point QQ. Watch out! Remember that the directional derivative depends on both the *gradient* of the function and the *direction vector*, so we need both!

STEP 2

1. Find the Gradient
2. Find the Direction Vector
3. Calculate the Dot Product

STEP 3

Let's **start** by calculating the gradient of our function f(x,y,z)=xy2z3f(x, y, z) = xy^2z^3.
Remember, the gradient is a vector that points in the direction of the greatest increase of the function.
It's like climbing a hill – the gradient tells you the steepest way up!

STEP 4

The gradient, denoted by ∇f\nabla f, is found by taking the partial derivatives of ff with respect to each variable.
So, let's **do it**! ∂f∂x=y2z3\frac{\partial f}{\partial x} = y^2z^3 ∂f∂y=2xyz3\frac{\partial f}{\partial y} = 2xyz^3∂f∂z=3xy2z2\frac{\partial f}{\partial z} = 3xy^2z^2

STEP 5

Therefore, our gradient is: ∇f=⟨y2z3,2xyz3,3xy2z2⟩\nabla f = \langle y^2z^3, 2xyz^3, 3xy^2z^2 \rangle

STEP 6

Now, we need to **evaluate** this gradient at our specific point P(2,1,1)P(2, 1, 1).
Plugging in x=2x = \textbf{2}, y=1y = \textbf{1}, and z=1z = \textbf{1}, we get: ∇f(2,1,1)=⟨(1)2(1)3,2(2)(1)(1)3,3(2)(1)2(1)2⟩=⟨1,4,6⟩\nabla f(2, 1, 1) = \langle (\textbf{1})^2(\textbf{1})^3, 2(\textbf{2})(\textbf{1})(\textbf{1})^3, 3(\textbf{2})(\textbf{1})^2(\textbf{1})^2 \rangle = \langle \textbf{1}, \textbf{4}, \textbf{6} \rangle So, the gradient of ff at point PP is ⟨1,4,6⟩\langle \textbf{1}, \textbf{4}, \textbf{6} \rangle.

STEP 7

We want the directional derivative in the direction from point P(2,1,1)P(2, 1, 1) to point Q(0,−3,5)Q(0, -3, 5).
To get this direction, we need to find the vector that goes from PP to QQ.

STEP 8

We can **find** this vector by subtracting the coordinates of PP from the coordinates of QQ: PQ⃗=⟨0−2,−3−1,5−1⟩=⟨-2,-4,4⟩\vec{PQ} = \langle 0 - 2, -3 - 1, 5 - 1 \rangle = \langle \textbf{-2}, \textbf{-4}, \textbf{4} \rangle

STEP 9

Now, we need a *unit vector* in this direction.
A unit vector has a length of 11, which makes it perfect for indicating direction without messing with the magnitude.

STEP 10

To **get** the unit vector, we divide our vector PQ⃗\vec{PQ} by its magnitude.
The magnitude of PQ⃗\vec{PQ} is: ∥PQ⃗∥=(−2)2+(−4)2+42=4+16+16=36=6\|\vec{PQ}\| = \sqrt{(-2)^2 + (-4)^2 + 4^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = \textbf{6}

STEP 11

So, our unit direction vector u⃗\vec{u} is: u⃗=PQ⃗∥PQ⃗∥=⟨−2,−4,4⟩6=⟨-1/3,-2/3,2/3⟩\vec{u} = \frac{\vec{PQ}}{\|\vec{PQ}\|} = \frac{\langle -2, -4, 4 \rangle}{\textbf{6}} = \langle \textbf{-1/3}, \textbf{-2/3}, \textbf{2/3} \rangle

STEP 12

Finally, the directional derivative is the dot product of the gradient at PP and our unit direction vector u⃗\vec{u}.
The dot product essentially tells us how much the gradient aligns with our chosen direction.

STEP 13

Let's **calculate** that dot product! ∇f(2,1,1)⋅u⃗=⟨1,4,6⟩⋅⟨−1/3,−2/3,2/3⟩=(1)(−1/3)+(4)(−2/3)+(6)(2/3)=−1/3−8/3+12/3=3/3=1\nabla f(2, 1, 1) \cdot \vec{u} = \langle 1, 4, 6 \rangle \cdot \langle -1/3, -2/3, 2/3 \rangle = (1)(-1/3) + (4)(-2/3) + (6)(2/3) = -1/3 - 8/3 + 12/3 = \textbf{3/3} = \textbf{1}

STEP 14

The directional derivative of f(x,y,z)f(x, y, z) at P(2,1,1)P(2, 1, 1) in the direction of Q(0,−3,5)Q(0, -3, 5) is **1**.

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