Let f be a function of two variables that has continuous...

Let $f$ be a function of two variables that has continuous partial derivatives and consider the points $A(1,3), B(3,3)$ $C(1,7),$ and $D(6,15) .$ The directional derivative of $f$ at $A$ in the direction of the vector $A B$ is 3 and the directional deriv- ative at $A$ in the direction of $\vec{A C}$ is $26 .$ Find the directional derivative of $f$ at $A$ in the direction of the vector $A D$

Let $f$ be a function of two variables that has continuous
partial derivatives and consider the points $A(1,3), B(3,3)$
$C(1,7),$ and $D(6,15) .$ The directional derivative of $f$ at $A$ in
the direction of the vector $A B$ is 3 and the directional deriv-
ative at $A$ in the direction of $\vec{A C}$ is $26 .$ Find the directional
derivative of $f$ at $A$ in the direction of the vector $A D$

Key Concepts

Directional Derivative

The directional derivative of a function provides the rate at which the function changes at a point in a specified direction. It is computed by taking the dot product of the gradient vector of the function and a unit vector that points in the chosen direction, thereby capturing both the magnitude and the direction of the function's steepest rate of change.

Gradient

The gradient of a function of several variables is a vector composed of its partial derivatives. This vector points in the direction of the steepest increase of the function and its magnitude gives the rate of maximum ascent. It is a fundamental concept in multivariable calculus for understanding how functions change in different directions.

Unit Vectors

Unit vectors are vectors that have a magnitude of one and are used to indicate direction. When computing directional derivatives, it is important to use a unit vector to ensure that the derivative reflects the rate of change per unit length in that direction.

Dot Product

The dot product is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. In the context of directional derivatives, it is used to combine the gradient vector and the unit direction vector, effectively projecting the gradient onto the direction of interest.

Transcript

00:01 In this problem, we are given the four points shown, and first told that if we take the directional derivative of f at the point 1 3, or a, in the direction of the vector ab, the directional derivative is going to be 3.

00:19 And so if we set that up, we first need what the directional vector is.

00:23 We don't know what the function f is, and so we need to eventually find the gradient vector.

00:29 But for now we need to find a directional vector.

00:33 In order to do that, we're going to find the vector ab by taking the point b and subtracting the point a, because b is our endpoint.

00:40 So we're going to take 3 -3 and subtract 1 -3, and our resulting vector is going to be 2 comma 0.

00:48 Now we have to make sure the specter's length is 1.

00:50 So we're going to divide it by the square root of the sum of each of the component squared, and so 2 squared plus 0 squared, 2 squared is 4, 0 squared is 0.

01:02 So the square root of 4 is 2.

01:04 So we're going to divide each the components by 2, and we get 1 ,0.

01:10 And so when we set up the equation for the first directional derivative, we're going to get some gradient vector, the partial of f with respect to x and the partial with respect to y.

01:23 Dot product with our directional vector, 1 comma 0, and that is equal to 3.

01:30 Now if we do this, dot product, we see that f of x times 1 .1.

01:35 Is just f of x plus f of y times zero that's just zero equals three so we know what f of the partial of f with respect to x is equal to now we're going to take the next information we were given which is that the directional derivative in the direction of the vector ac is equal to 26 and so to find the directional vector for ac we're going to take the point c which is 1 comma 7 subtract the point a which is 1 comma 3 and we do that we get the vector for a vector 0 comma 4.

02:11 And we're going to make the length of this vector 1 by dividing it by the square root of the sum of each component squared, which is going to be 0 squared plus 4 squared.

02:22 0 squared is 0.

02:23 4 squared is 16.

02:24 The square root of 16 is 4.

02:25 Distribute that into each component, we get 0, comma 1.

02:29 Now we're going to set up the directional derivative equation again, where we get some partial with respect to x of f, partial with respect to y...

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