<html> <p><img src="https://i.stack.imgur.com/VOPuE.png" width="466" height="475"/></p> <p><a href="https://mathematica.stackexchange.com/questions/27352/adding-tangent-velocity-vectors-to-a-plot-of-a-space-curve">Source</a></p> <p><br></p> <h2>Introduction</h2> <p> Hello it's a me again drifter1!</p> <p> Today, we will continue with our <strong>Mathematical Analysis</strong> series of Mathematics by getting into <strong>Directional Derivatives</strong> that are based on <a href="https://steemit.com/mathematics/@drifter1/mathematics-mathematical-analysis-partial-derivatives">Partial Derivatives</a> that we covered last time.</p> <p> As I already said last time, Directional Derivatives are used for <strong>finding normal vectors to a plane </strong>and also for <strong>finding tangent planes</strong>.</p> <p>There is not much more to say about, so let's get started!</p> <h2>Small recap</h2> <p>Last time we talked about Derivatives of vector functions (curves) on top of planes.</p> <p>If z = f(x, y) and r(t) = x(t)i + y(t)j then the <strong>derivative </strong><em><strong>along the</strong></em><strong> curve</strong> is given by:</p> <p><em>f '(t) = df/dt = ∂f/∂x * dx/dt + ∂f/∂y * dy/dt</em></p> <p>We can see that we use the <strong>rule of composition</strong>, because those functions are "complex".</p> <p> The function f(x, y) is a function of x and y, but x(t) and y(t) are functions of t. This means that we first partially differentiate f(x, y) by x or y and multiply this with the derivative of x(t) or y(t) with t.</p> <p>Of course the derivative along a curve can also be found for 3 or even more dimensions...</p> <p><br></p> <h2>Tangent vectors</h2> <p>Suppose the plane f(x, y, z) = c and the curve r(t) = x(t)i + y(t)j + z(t)k.</p> <p>The derivative along the curve is:</p> <p><em>u = df/dt = ∂f/∂x * dx/dt + ∂f/∂y * dy/dt + ∂f/∂z* dz/dt </em></p> <p><br></p> <p>The <strong>tangent vector of the curve r</strong> is:</p> <p><img src="http://quicklatex.com/cache3/99/ql_060b20d8dd26b1568a29e901c86a3e99_l3.png"/></p> <p><a href="http://quicklatex.com/cache3/99/ql_060b20d8dd26b1568a29e901c86a3e99_l3.png">quicklatex</a></p> <p><br></p> <p>The <strong>gradient of f</strong> (or Del of f) is the vector:</p> <p> ∇<em>f = gradf = (∂f/∂x, ∂f/∂y, ∂f/∂z)</em></p> <p><br></p> <p>So, the <strong>derivative along the curve</strong> is:</p> <p><em>df/dt = (∂f/∂x, ∂f/∂y, ∂f/∂z) * (dx/dt, dy/dt, dz/dt) = </em>∇<em>f * u</em></p> <p>where ∇<em>f is the gradient of f and u is a tangent vector of curve r(t) = (x(t), y(t), z(t))</em></p> <p><br></p> <p>If u =(a, b, c) is a unit vector (|u| = 1) and P0 a Point in space then:</p> <p>∇<em>f(P0) * u = Duf(P0)</em></p> <p>is called the <strong>derivative of f </strong><em><strong>along </strong></em><strong>the direction of u = (a, b, c)</strong>.</p> <p><br></p> <p>For <strong>example</strong>:</p> <p><img src="http://quicklatex.com/cache3/2a/ql_6f4ca90c71aa8c0efd9f45a763ad0b2a_l3.png"/></p> <p><img src="http://quicklatex.com/cache3/da/ql_5f3f908f3bf0a30edc51683b467e13da_l3.png"/></p> <p><img src="http://quicklatex.com/cache3/f8/ql_6a63aae5ef346d3257a09da9d05a98f8_l3.png"/></p> <p>Duf(P) = ?</p> <p>The given vector u is not an unit vector and so we have to <em>convert </em>it.</p> <p><img src="http://quicklatex.com/cache3/02/ql_2214cd126f9df59305271d33c303d902_l3.png"/></p> <p><img src="http://quicklatex.com/cache3/43/ql_9414dd19ade221fb6dcd63d2f68dec43_l3.png"/></p> <p>The gradient of f is:</p> <p>∇f = (∂f/∂x, ∂f/∂y) = (2x + 2y, 2x - 6y)</p> <p>For the point P we have:</p> <p>∇f(P) = (1 + 1, 1 - 3) = (2, -2)</p> <p>That way:</p> <p>Duf(P0) = ∇f(P) * u0 =></p> <p><img src="http://quicklatex.com/cache3/09/ql_5d33e630ef6e4296eb519db51a356509_l3.png"/></p> <p>Some equations through <a href="http://quicklatex.com/">quicklatex.</a></p> <p><br></p> <h2>An explanation</h2> <p> Suppose we have the same plane and curve as before, the unit vector u = (u1, u2, u3) and the Point P0(x0, y0, z0).</p> <p>And ε is a line that passes through P0, is parallel to u and has the parametric equations:</p> <p><em>x = x0 + u1t</em></p> <p><em>y = y0 + u2t</em></p> <p><em>z = z0 + u3t</em></p> <p>Then dx/dt = u1, dy/dt = u2, dz/dt = u3 and so:</p> <p>df/dt = ∇f(P0) * (u1, u2, u3) = ∇f(P0) * u = Duf(P0) </p> <p>Which of course is the derivative of f along the direction of u.</p> <p><br></p> <h2>Some interesting cases</h2> <p>We can write the same equation like that:</p> <p><em>Duf(P0) = ∇f(P0) * u = |∇f(P0) ||u|*cosφ</em></p> <p>where φ is the angle between the vectors ∇f(P0) and u.</p> <blockquote>This is clearly a "dot" product between vectors...</blockquote> <p><br></p> <p>Depending on the angle we have the following <strong>cases</strong>:</p> <ol> <li>φ = 0, ∇f(P0) and u are parallel => Duf(P0) = |∇f(P0)| (highest possible increase of f)</li> <li>φ = π, ∇f(P0) = -u => Duf(P0) = -|∇f(P0)| (highest possible decrease of f)</li> <li>φ = π/2, ∇f(P0) and u are across and so Duf(P0) = 0</li> </ol> <p>And so:</p> <p><em>-|∇f(P0)| <= Duf(P0) <= |∇f(P0)|</em></p> <p><br></p> <h2>Normal vectors to a plane</h2> <p>So, now after all that we can finally get into the actual useful thing of this post...</p> <p> Suppose S: f(x, y , z) = c being a plane/surface and r(t) = x(t)i + y(t)j +z(t)k a curve on top of it that passes through the point P0(x0, y0, z0).</p> <p>u = dr/dt = (dx/dt, dy/dt, dz/dt) is a tangent vector of curve r at P0.</p> <p><br></p> <p>Let's also suppose N = (a, b, c) being a <strong>normal vector to our plane</strong>.</p> <p>Of course we can find a ∇f(P0) so that:</p> <p>∇f(P0) * u = 0, which means that they are vertically across..</p> <p>So, <strong>N = ∇f(P0) </strong></p> <p><br></p> <h2>Tangent planes</h2> <p>By using this concept, planes/surfaces of the form:</p> <p><em>a(x - x0) + b(y- y0) + c(z - z0) = 0</em></p> <p>have N = (a, b, c) as a normal vector and also pass through P0.</p> <p><br></p> <p>So, with f(x, y, z) = c being a plane and P0(x0, y0, z0) a point,</p> <p>the <strong>equation of a tangent plane</strong> is:</p> <p><em>∂f/∂x (x - x0) + ∂f/∂y (y - y0) + ∂f/∂z (z - z0) = 0</em></p> <p><br></p> <p>For a surface f(x, y) = z we have:</p> <p>F(x, y, z) = f(x, y) - z = 0</p> <p>And so;</p> <p><em>∂F/∂x = ∂f/∂x</em></p> <p><em>∂F/∂y = ∂f/∂y</em></p> <p><em>∂F/∂z = -1</em></p> <p>Which means that:</p> <p>N = ∇f(P0) = (<em>∂f/∂x, ∂f/∂y, -1</em>)</p> <p>And finally the equation of such a tangent plane is:</p> <p><em>∂f/∂x (x - x0) + ∂f/∂y (y - y0) - (z - z0) = 0</em></p> <p><br></p> <p><br></p> <p>And this is actually it for today and I hope that you enjoyed it!</p> <p>Next time we will start getting into double/surface integrals!</p> <p>Bye!</p> </html>
author | drifter1 |
---|---|
permlink | mathematics-mathematical-analysis-directional-derivatives |
category | mathematics |
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Hey, just wanted to let you know I gave you an upvote because I appreciate your content! =D See you around
author | greentomorrow |
---|---|
permlink | re-drifter1-mathematics-mathematical-analysis-directional-derivatives-20180418t211422995z |
category | mathematics |
json_metadata | {} |
created | 2018-04-18 21:14:21 |
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