13.6 Directional derivatives,

Transcription

1 13.5 The chain rule Theorem [Chain Rule for Functions of Two Variables] If w = f ( x, y ) is differentiable and x and y are differentiable functions of t, then w is a differentiable function of t and dw = dt x f ( xy, ) dx dt + y f ( xy, ) dy dt. Theorem [Chain Rule for Functions of Three Variables] If w = f ( xyz,, ) is differentiable and x, y and z are differentiable functions of t, then w is a differentiable function of t and dw = dt x f ( xyz,, ) dx dt + y f ( xyz,, ) dy dt + z f ( x, y, z) dz dt. Theorem [Chain Rule for Two Independent Variables and Three Intermediate Variables] If w = f ( x, y, z ), x = g ( rs,, ) y = h ( r, s ) and z = k ( rs., ) If all four functions are differentiable, then w has partial derivatives with respect to r and s, given by the formulas r w= x w r x + y w r y + z w r z s w= x w s x + y w s y + z w s z. Theorem [A Formula for Implicit Differentiation] Suppose F ( xy, ) is differentiable and the equation F ( xy, ) = 0 defines y implicitly as a differentiable function of x. Then, at any dy point where F y 0, = F x. dx F y 13.6 Directional derivatives,

2 gradient vectors, and tangent planes Definition [Directional Derivatives] The derivative of f at P 0 (, y 0 ) in the direction of the unit vector u=u 1 i+u 2 j is the number df f ( x 0 + su 1, y 0 + su 2 ) f ( x 0, y 0 ) =lim provided the limit exists. ds s 0 s u, P 0 x 0 Definition [Gradient Vector or Gradient] The gradient vector (gradient) of f ( xy, ) at a point P 0 ( x 0, y 0 ) is the vector grad( f ) = x fi+ fj obtained by evaluating the y partial derivatives of f at P 0. Theorem [The Directional Derivative is a Scalar Product] If the partial derivatives of f ( x, y ) are defined at P 0 (, y 0 ), then x 0 df = grad( f ).u, the scalar product of the gradient f at P ds P0 0 and u. u, P 0 Properties of the Directional Derivative D u f = grad( f ).u= grad( f ) cos( θ ) where θ is the angle between grad( f ) and u. The function f increases most rapidly when cos( θ) = 1 or when u is the direction of grad( f ). That is, at each point P in its domain, f increases most rapidly in the direction of the gradient vector grad( f ) at P. The derivative in this direction is D u f = grad( f ).u= grad( f ) cos( 0 ) = grad( f ). Similarly, f decreases most rapidly in the direction - grad( f ). The derivative in this direction is D u f = grad( f ).u= grad( f ) cos( π ) =- grad( f ).

3 Any direction u orthogonal to the gradient is a direction of zero change in f because θ then equals π 2 and D u f = grad( f ).u= grad( f ) cos π 2 = grad ( f ).0=0. Definition [Tangent Plane and Normal Line] The tangent plane at the point P 0 ( x 0, y 0, z 0 ) on the level surface f ( xyz,, ) = c is the plane through P 0 normal to grad( f ). The normal line of the surface at P P0 0 is the line through P 0 parallel to grad( f ). P0 Plane tangent to a surface z = f ( xy, ) at ( x 0, y 0, f ( x 0, y 0 )) The plane tangent to the surface z = f ( xy, ) at the point P 0 ( x 0, y 0, z 0 ) =( x 0, y 0, f ( x 0, y 0 )) is f x ( x 0, y 0 )( x x 0 ) + f y ( x 0, y 0 )( y y 0 ) = 0. Estimating the change in f in a direction u To estimate the change in the value of a function f when we move a small distance ds from a point P 0 in a praticular direction u, use the formula df = grad( f).uds. P0 Algebra rules for gradients Constant multiple rule: grad( kf ) =k grad( f ) (any number k) Sum Rule: grad ( f + g ) = grad( f ) + grad( g ) Difference Rule: grad ( f g ) = grad( f ) grad( g ) Product Rule: grad( fg ) = f grad( g) + g grad( f ) Quotient Rule: grad f g grad( f) f grad( g ) = g g 2 A First Look at Tangent Planes Worksheet by Mike May, S.J.- maymk@slu.edu

4 A first naive way to look at tangent planes is to take the tangent lines to the two cross section curves at a point and turn those lines into a plane. (The naive approach works if the function is differentiable, but that is a detail we will look at later.) This worksheet is intended as a demonstration of that technique. Constructing a plane defined by tangent lines to cross sections We start by defining a function that we will work with. Let z=f(x,y) =x^2-3*x*y+y^2. It is first of all useful to look at the graph. f:= (x,y) - x^2-3*x*y+y^2; f := ( xy, ) x 2 3 xy+ y 2 plot3d( f(x,y), x=-5..5, y=-5..5, axes=framed); Notice that the graph goes up in one direction and down in another. We are at a saddle, so tangent planes will tend to cut the surface, much as the tangent line to a cubic curve cuts the curve at the inflection point. To find the tangent plane we considered cross sections. We will start with the point (-1,3). We start by finding the z-value at the point on the graph. Next compute the functions in one variable that we obtain by holding either x or y constant. Now compute the derivatives of the functions in one variable and substitute in the point to find an x-slope and y-slope. graph := plot3d({f(x,y)}, x=0..5, y=0..5, color=blue): tanplane :=plot3d({19-11*(x+1)+9*(y-3)}, x=0..5, y=0..5, color=green):

5 plots[display](graph, tanplane);?plots,spacecurve plot3d(f(x,y),x= , y= , view= , axes=normal, color=blue);

6 x0 := -1; y0 := 3; z0 := f(x0, y0); x0 := -1 y0 := 3 z0 := 19 fx := f(x,y0); fy := f(x0,y); fx := x 2 9 x + 9 fy := y+ y 2 Dfx := diff(fx,x); mx := subs(x=x0,dfx); Dfy := diff(fy,y); my := subs(y=y0,dfy); Dfx := 2 x 9 mx := -11 Dfy := y my := 9 We see that the x slope is -11, the y-slope is 9, and the z value is 19. Let's try plotting the surface and the plane we have obtained. The two surfaces look like good approximations to each other. Let us clean up the picture with some more magical Maple code. Since we want to plot the cross-sections of the graph, we use the command spacecurve in the plots package with(plots): s1 := plot3d(f(x,y), x=-3..1, y=1..5, view=0..40, color=blue): s2 := plot3d(19-11*(x+1)+9*(y-3), x=-3..1, y=1..5, view=0..40, color=red): c1 := spacecurve([x,y0,f(x,y0)], x=-3..1, color=green, thickness =3): c2 := spacecurve([x0,y,f(x0,y)], y=1..5, color=yellow, thickness =3): display3d({s1,s2, c1, c2}, axes=boxed); Warning, the name changecoords has been redefined

7 Thus we see that we have constructed a point that contains the appropriate point of the surface and contains the two tangent lines. An automated approach We can set up a block of code that does the work of the example all in one step. f := (x,y) - x^2-3*x*y+y^2; x0:=-1; y0:=3; width := 5: z0:=f(x0,y0); fx:=f(x,y0); Dfx := diff(fx,x); xslope:= subs(x=x0,dfx); fy:=f(x0,y); Dfy := diff(fy,y); yslope:= subs(y=y0,dfy); fsurface := plot3d(f(x,y), x=x0-width..x0+width, y=y0-width..y0+width, color=red): ftanplane := plot3d(z0+xslope*(x-x0)+yslope*(y-y0), x=x0-width..x0+width, y=y0-width..y0+width, color=blue): xcurve := spacecurve([x,y0,f(x,y0)], x=x0-width..x0+width,

8 color=green, thickness =3): ycurve := spacecurve([x0,y,f(x0,y)], y=y0-width..y0+width, color=yellow, thickness =3): display3d({fsurface,ftanplane, xcurve, ycurve}, axes=boxed); f := ( xy, ) x 2 3 xy+ y 2 x0 := -1 y0 := 3 z0 := 19 fx := x 2 9 x + 9 Dfx := 2 x 9 xslope := -11 fy := y+ y 2 Dfy := y yslope := 9 The advantage of this set-up is that we can consider a different example by modifying the first 2 lines of the block of code above and re-executing the block of code. Exercises 1) Modify the code above to examine the function x xy+ 6 y 2 at two points of your

9 choosing. Verify that the plane obtained is tangent to the surface. (You may want to reduce the width variable to zoom in on the point of tangency.) 2) Modify the code above to examine the function 3 x x 3 + y 3 3 y at the points {(-1, -1), (-1, 1), (1, -1), (1,1)}. Explain what you find. Visualizing Directional Derivatives and Gradients Worksheet by Mike May, S.J.- maymk@slu.edu As we have been looking at directional dericatives and gradients, it seems worthwhile to look at a Maple visualization of everything we have been doing. First want to load the appropriate commands. We will need commands from the packages linalg, plottools, and plots. (Restart cleans up in case we were already running a Maple worksheet.) restart: with(linalg): with(plottools): with(plots): Warning, the protected names norm and trace have been redefined and unprotected Warning, the names arrow and changecoords have been redefined Directional derivatives: For preliminaries we want to define the function, the point in the x-y plane we are evaluating at and the direction vector we are interested in taking a derivative in.. f := (x,y) - x^2 + 2*y^2; xval :=2: yval:= 3: dirvec := [3, 4]: f := ( xy, ) x y 2 Next we would like to compute a unit vector in the desired direction and the z-coordinate of the point of interest. uvec := convert( evalm(dirvec/sqrt(dotprod(dirvec,dirvec))), list); zval := f(xval, yval); 3 4 uvec :=, 5 5 zval := 22 To visualize what we are doing, we look at the surface together with a plane containing the direction vector. surf:= plot3d(f(x,y), x=xval-2..xval+2, y=yval-2..yval+2, axes=boxed, color=red, style = patchnogrid):

10 pathc :=spacecurve( [xval+t*uvec[1], yval+t*uvec[2], f(xval+t*uvec[1], yval+t*uvec[2])], t=-2..2, color=blue, thickness=2): cutplane := polygon([[xval+2*uvec[1], yval+2*uvec[2],zval+5], [xval+2*uvec[1], yval+2*uvec[2],zval-5], [xval-2*uvec[1], yval-2*uvec[2],zval-5], [xval-2*uvec[1], yval-2*uvec[2],zval+5]], color=yellow): display3d({cutplane, surf, pathc}); The surface is in red, the plane that cuts the cross section in the desired direction is in yellow,

11 and the curve in the cross section is in blue. We can get a better view if we look at the function as a function of one variable in the cutting plane. plot(f(xval+t*uvec[1],yval+t*uvec[2]), t=-2..2); We can compute the directional derivative from the definition. dderiv := limit((f(xval+t*uvec[1], yval+t*uvec[2]) - f(xval, yval))/t, t=0); dderiv := 12 This can be used to plot a tangent line to the curve. plot({f(xval+t*uvec[1],yval+t*uvec[2]), zval+dderiv*t}, t=-2..2);

12 We can also put this line back on the 3D graph of the surface. tanline :=spacecurve([xval+t*uvec[1], yval+t*uvec[2], zval+t*dderiv], t=-2..2, color=green, thickness=3): display3d({cutplane, surf, pathc, tanline});

13 For convenience sake we can put the stuff together into one block of code. f := (x,y) - x^2 + 2*y^2; xval :=2: yval:= 3: dirvec := [3, 4]: uvec := convert( evalm(dirvec/sqrt(dotprod(dirvec,dirvec))),list); zval := f(xval,yval); dderiv := limit((f(xval+t*uvec[1], yval+t*uvec[2]) - f(xval, yval))/t, t=0); plot({f(xval+t*uvec[1],yval+t*uvec[2]), zval+dderiv*t}, t=-2..2); surf:= plot3d(f(x,y), x=xval-2..xval+2, y=yval-2..yval+2, axes=boxed, color=red, style = patchnogrid): pathc :=spacecurve([xval+t*uvec[1], yval+t*uvec[2], f(xval+t*uvec[1], yval+t*uvec[2])], t=-2..2, color=blue, thickness=2): cutplane := polygon([[xval+2*uvec[1], yval+2*uvec[2],zval+5], [xval+2*uvec[1], yval+2*uvec[2],zval-5], [xval-2*uvec[1], yval-2*uvec[2],zval-5], [xval-2*uvec[1], yval-2*uvec[2],zval+5]], color=yellow):

14 tanline :=spacecurve([xval+t*uvec[1], yval+t*uvec[2], zval+t*dderiv], t=-2..2, color=green, thickness=3): display3d({cutplane, surf, pathc, tanline}); f := ( xy, ) x y uvec :=, 5 5 zval := 22 dderiv := 12

15 Exercise:. 1) Let g ( xy, ) = sin ( + y 2 ). Modify the code above to plot the graph of g(x,y) in a region around the point (3, 4) together with a line tangent in the direction (5, 12). x 2 A second look at directional derivatives The alternate way to produce the directional derivative is to compute the partial derivatives, evaluate them at our point, and then take the sum of the partials times the appropriate component of the unit direction vector. f := (x,y) - x^2 + 2*y^2: xval :=2: yval:= 3: dirvec := [3, 4]: fx := diff(f(x,y),x); xslope := subs({x=xval, y=yval}, fx); fy := diff(f(x,y), y); yslope := subs({x=xval, y=yval}, fy); dirderiv := xslope*uvec[1]+yslope*uvec[2]; fx := 2 x xslope := 4 fy := 4 y yslope := 12

16 dirderiv := 12 Notice that we do get the same answer for the directional derivative that we got with the formal definition above. This second method is more roundabout, but it has the advantage of being easier to do when we don't have Maple handy. In the second method we only have to take limits for two cross sections. Exercise:. 2) Let g ( xy, ) = sin ( x 2 + y 2 ). Find the directional derivative of g(x,y) at the point (3, 4) in the direction (5, 12) by the method that uses the partials at the point. Comparing the two methods of finding directionalderivatives It is worthwhile varifying that the two methods of computing derectional derivative give the same answer for any direction. (To be more precise, that these two methods with produce the same answer in every direction is the definition of the function being differentiable at a point.) We recall computations made above to get the x-slope and y-slope at a point. f := (x,y) - x^2 + 2*y^2: xval :=2: yval:= 3: dirvec := [3, 4]: zval := f(xval,yval); fx := diff(f(x,y),x): xslope := subs({x=xval, y=yval},fx): fy := diff(f(x,y), y): yslope := subs({x=xval, y=yval}, fy): zval := 22 We recall that a unit vector in direction theta has coordinates (cos(theta), sin(theta)). We then find a directional derivative in direction theta by using the definition of directional derivative. ddera := theta - limit((f(xval+t*cos(theta), yval+t*sin(theta))-f(xval, yval))/t, t=0); ddera := θ lim t 0 f ( xval + t cos( θ ), yval + t sin( θ )) f ( xval, yval) t We now compute the directional derivative by the method that uses x-slope and y-slope. dderb := theta - cos(theta)*xslope + sin(theta)*yslope; dderb := θ cos( θ) xslope + sin( θ) yslope It is now a simple matter to graphically compare the values obtained by the two methods for all angles theta. plot({ddera(theta)- dderb(theta)}, theta=0..2*pi, axes=boxed);

17 If we plot the ring of points (a+cos(theta), b+sin(theta), f(a,b)+ddera(theta)) we get a ring in the tangent plane. surf:= plot3d(f(x,y), x=xval-2..xval+2, y=yval-2..yval+2, axes=boxed, color=red, style = patchnogrid): tanplane:= plot3d(zval+xslope*(x-xval)+yslope*(y-yval), x=xval-2..xval+2, y=yval-2..yval+2, axes=boxed, color=yellow, style = patchnogrid): tanring :=spacecurve([xval+cos(theta), yval+sin(theta), f(xval,yval)+ddera(theta)], theta=0..2*pi, color=blue, thickness=2): display3d({surf, tanplane, tanring});

18 Exercise:. 3) Let g ( xy, ) = sin ( x 2 + y 2 ). Show that g(x,y) is differentiable at (3, 4) by showing that the 2 methods of computing directional derivatives give the same answer in all directions. Gradients: The gradient of a function can be thought of in a number of ways. The easiest description is symbolic. The gradient of f(x,y) is defined to be the vector with components equal to the partial derivatives. In the terminology we have been using, grad(f) = [x-slope, y-slope]. Notice that the gradient has as many components as the input vector, rather than the number of coordiantes in a point in the graph. f := (x,y) - x^2 + 2*y^2; xval :=2: yval:= 3: fx := diff(f(x,y),x); xslope := subs({x=xval, y=yval},fx); fy := diff(f(x,y), y); yslope := subs({x=xval, y=yval}, fy); gradient := [xslope,yslope]; f := ( xy, ) x y 2 fx := 2 x xslope := 4

19 fy := 4 y yslope := 12 gradient := [ 412, ] We are also told that the gradient should be the direction that is perpendicular to the contour of the graph at that point. To see that we want to put the contourplot and the gradient on the same graph. cplot := contourplot(f(x,y),x=xval-2..xval+2, y=yval-2..yval+2): gradv := line([xval,yval], [xval+xslope, yval+yslope], linestyle=1, thickness=3, color=green): display({cplot, gradv},scaling=constrained); This relationship can be seen more clearly with the Maple command gradplot that plots a field of scaled gradient vectors. gplot := gradplot(f(x,y),x=xval-2..xval+2, y=yval-2..yval+2): cplot := contourplot(f(x,y),x=xval-2..xval+2, y=yval-2..yval+2): display({cplot, gplot}, scaling=constrained);

20 Tied into the idea that the gradient is perpendicular to the contours is that it is the direction of steepest ascent. This means that we can visually check to see if we have computed the gradient correctly. If we take the cross section in the direction of the gradient it should be the direction that is rising most steeply. f := (x,y) - x^2 + 2*y^2; xval :=2: yval:= 3: zval := f(xval,yval): fx := diff(f(x,y),x): mx := subs({x=xval, y=yval},fx): fy := diff(f(x,y), y): my := subs({x=xval, y=yval}, fy): gradvec := [mx, my]; unitgradient := convert(evalm(gradvec/sqrt(dotprod(gradvec,gradvec))),list); gradientderiv := mx*unitgradient[1]+my*unitgradient[2]; surf:= plot3d(f(x,y), x=xval-2..xval+2, y=yval-2..yval+2, axes=boxed, color=red, style = patchnogrid): gradientpath :=spacecurve([xval+t*unitgradient[1], yval+t*unitgradient[2], f(xval+t*unitgradient[1], yval+t*unitgradient[2])], t=-2..2, color=blue, thickness=2):

21 cutplane := polygon([[xval+2*unitgradient[1], yval+2*unitgradient[2],zval+5], [xval+2*unitgradient[1], yval+2*unitgradient[2],zval-5], [xval-2*unitgradient[1], yval-2*unitgradient[2],zval-5], [xval-2*unitgradient[1], yval-2*unitgradient[2],zval+5]], color=yellow): gradline :=spacecurve( [xval+t*unitgradient[1], yval+t*unitgradient[2], zval+t*gradientderiv], t=0..1, color=green, thickness=3): display3d({cutplane, surf, gradientpath, gradline}); f := ( xy, ) x y 2 gradvec := [ 412, ] 1 unitgradient := 10, gradientderiv := Exercise: 4a) Let g ( xy, ) = x x + 3 y 2 4 y. Estimate the gradient of g(x,y) at (3, 4) by plotting a contouplot and finding the direction perpendicular to the contours.

22 4b) Estimate the gradient of g(x,y) at (3, 4) by plotting the surface for a region around (3,4) and estimating the direction of steepest increase. 4c) Calculate the gradient of g(x,y) at (3, 4) symbolically.