Speaker A
The topic of this video is computing directional derivatives with gradients. Let F be a differentiable function at the point AB, and let U, which equals the vector U sub one, U sub two, be a unit vector in the XY plane.
Learn how to compute directional derivatives using gradients with geometric interpretation and a detailed example.
Key Takeaways
- Directional derivatives measure the rate of change of a function in any direction.
- The gradient vector points in the direction of the steepest ascent of the function.
- The directional derivative is computed as the dot product of the gradient and a unit direction vector.
- Vectors must be normalized to unit vectors before computing directional derivatives.
- Geometric visualization helps in understanding the meaning of directional derivatives on surfaces.
What the video covers
- Introduction to directional derivatives and gradients for differentiable functions at a point.
- Definition of directional derivative as the dot product of the gradient and a unit vector.
- Geometric interpretation involving a surface, a point, a vector, and tangent lines to curves.
- Explanation of how the directional derivative represents the slope of a curve on the surface.
- Step-by-step example using the function F(x,y) = x sin(y) at the point (1, π).
- Calculation of the gradient vector components via partial derivatives.
- Evaluation of the gradient vector at the specific point (1, π).
- Conversion of a given vector into a unit vector by dividing by its magnitude.
- Computation of the directional derivative as the dot product of the gradient and the unit vector.
- Final result of the directional derivative for the example is -4/5.
Chapters
- 00:00Introduction to directional derivatives and gradients
- 00:18Definition of directional derivative using dot product
- 00:40Geometric meaning of directional derivatives
- 00:57Constructing line and plane for geometric interpretation
- 01:37Slope of tangent line as directional derivative
- 01:54Example: function F(x,y) = x sin(y) at (1, π)
- 02:12Calculating gradient vector components
- 02:41Evaluating gradient at point (1, π)
- 03:01Converting given vector to unit vector
- 04:03Computing directional derivative using dot product
Full Transcript — Download SRT & Markdown
Speaker A
The directional derivative of F at the point AB in the direction of the vector U is denoted by this notation, and it equals the dot product of the gradient of F evaluated at AB and U. Before we apply this formula to an
Speaker A
example, let's review the geometric meaning of the directional derivative. Here's a surface Z equal F of XY and a point AB in the XY plane, the corresponding point AB F of AB on the surface, and the unit vector U, which
Speaker A
equals the vector U1, U2. To explain the meaning of the directional derivative of F at AB in the direction of U, let's first draw a line L passing through the point AB parallel to the vector U. Then we place a vertical plane in this figure
Speaker A
passing through the line L. Then observe that this plane and the surface intersect in a curve. Finally, we sketch a line in the plane that's tangent to the curve at the point AB F of AB. The slope of this line is the
Speaker A
directional derivative of F at AB in the direction of U. So when we calculate the directional derivative of F at the point AB in the direction of U, we are calculating the slope of a curve on the surface at the point AB F of AB in the
Speaker A
direction of the vector U. So let's compute the directional derivative of the function F of XY = X sin of Y at the point 1, Pi in the direction of the vector 3, -4. In this case, the directional
Speaker A
derivative is equal to the dot product of the gradient of F evaluated at 1, Pi and the vector U. So let's begin by calculating the gradient of F. This is a vector whose first component is the partial derivative of F with respect to
Speaker A
X, which equals sin of Y, and the second component of the vector is the partial derivative of F with respect to Y, which would be X cosine of Y. Now we want to calculate the value of this gradient
Speaker A
vector at the point 1, Pi. So we replace X with 1 and Y with Pi in this vector to obtain the vector sine of Pi, 1 times the cosine of Pi, which equals 0, 1. To calculate the directional derivative, we also need to know the
Speaker A
unit vector U, and the problem here is that the vector we're given, 3, -4, is not a unit vector. So let's call this vector V. What we need is a unit vector U that's parallel to V, that has the same
Speaker A
direction as V. And recall that U equals V divided by the magnitude of V, which is equal to the vector 3, -4 over the square root of 3 squared plus -4 squared. And since the denominator simplifies, there is the square root of 25 or 5. We have 3, -4 over 5,
Speaker A
which equals 3 fifths, -4 fifths. So we know the gradient vector at 1, Pi equals 0, 1, and we know the unit vector is 3 fifths, -4 fifths. Therefore, the directional derivative of F at 1, Pi in the direction of U equals the gradient of F evaluated
Speaker A
at 1, Pi dotted with the unit vector U, which equals 0, 1 dotted with 3 fifths, -4 fifths, which equals -4 fifths.
Topics:directional derivativegradientmultivariable calculuspartial derivativesvector calculusunit vectordot productgeometric interpretationfunction slopecalculus example
![[EN] How To Make Your Bigger And Stronger‼ everything you need to know — Transcript](https://i.ytimg.com/vi/q6wOgvbfJwM/maxresdefault.jpg)
