ENGI 344 4 Partial Differentiation Page 4-0 4. Partial Differentiation For functions of one variable, be found unambiguously by differentiation: y f x, the rate of change of the dependent variable can dy f x. In this chapter we explore dx f x, y. rates of change for functions of more than one variable, such as Contents: 4. Partial Derivatives - introduction, chain rule, practice 4. Higher Partial Derivatives, Clairaut s theorem, Laplace s PDE 4.3 Differentials; error estimation; chain rule [again]; implicit functions; partial derivatives on curves of intersection 4.4 The Jacobian - implicit and explicit forms; plane polar; spherical polar 4.5 Gradient Vector, directional derivative, potential function, central force law f x, y 4.6 Extrema; Second Derivative Test for 4.7 Lagrange Multipliers; nearest point on curve of intersection to given point 4.8 Miscellaneous Additional Examples
ENGI 344 4. Partial Derivatives Page 4-0 4. Partial Derivatives - introduction, chain rule, practice Example 4.. At a particular instant, a cone has a height of h = m and a base radius of r = m. The base radius is increasing at a rate of mm/s. The height is constant. How fast is the volume V increasing at this time? V 3 r h We need dv dt. h = const. V is a function of r only. dv dv dr dt dr dt chain rule d dr dr 3 dr dt 3 dt r h rh 3 000 3 m s Therefore dv dt 3 3 m s 0.0049 m s 750 dr dh dt dt dv how do we find? dt We shall return to this question later. But if mms and mms,
ENGI 344 4. Partial Derivatives Page 4-03 Graph of V against r and h : Plotting = V (where V = r h / 3) against both x = r and y = h yields The cross-section of this surface in the vertical plane h = is The cross-section of this surface in the vertical plane r = is V r r 0 r 0 4 3 r 0 lim,, V r r h V r h lim r V r 3 3 r r r r h r h lim r 0 r lim rh h r rh r 3 3 3 at h V h h 0 lim h 0,, V r h h V r h lim h r 3 V h
ENGI 344 4. Partial Derivatives Page 4-04 The tangent line to the surface in a cross-section (h = constant) has a slope of The tangent line to the surface in a cross-section (r = constant) has a slope of At each point on the surface, these two tangent lines define a tangent plane. V r. V h. V is a function of r and h, each of which in turn is a function of t only. In this case, the chain rule becomes In example 4.., dv V dr V dh dt r dt h dt V 4 V r =, h =, rh, r, r 3 3 h 3 3 dr dh dv 4, ms dt 000 dt 000 dt 3 000 3 000 500 3 Alternative notations: V Vr DrV V r, h r r If w = f (x, y, ), then w y,,,,, etc. y lim f x y y f x y y 0 A Maple worksheet, used to generate the graph of V r, h 3 r h "http://www.engr.mun.ca/~ggeorge/344/demos/conevolume.mws"., is available at Open this worksheet in Maple and click on the graph. Then, by dragging the mouse (with left button down), one can change the direction of view of the graph as one wishes. Other features of the graph may be changed upon opening a menu with a right mouse click on the graph or by using the main menu at the top of the Maple window.
ENGI 344 4. Partial Derivatives Page 4-05 Example 4..,, f x y x y x y Find f 0,3,4 / [the first partial derivative of f with respect to, evaluated at the point (0, 3, 4)]. d Apply the general differentiation rule dx f x y / x y 0 0 f f 4 4 f 0, 3, 4 0 9 6 5 n n f x n f x f x : OR f x y Using implicit differentiation, f f 0 0 f x y f f 4, as before 5 In this example, f = (distance of the point (x, y, ) from the origin).
ENGI 344 4. Partial Derivatives Page 4-06 Example 4..3 u x y / Find the three first partial derivatives, ux, uy, u. n n y/ x n x y d y y x x x x dx x y y/ y ux x yu x y/ y u x exp ln x y u uy ln x exp ln x ln x y / uy x ln x OR ln u y y ln x ln u ln x y y u uy ln x uy ln x u ln u y ln x u y ln x u u y y y / u ln x x ln x
ENGI 344 4. Higher Partial Derivatives Page 4-07 4. Higher Partial Derivatives, Clairaut s theorem, Laplace s PDE u x u x u x x u x u xx u u y x y x u xy 3 u u yx yx u x y u y y x u y x 3 u x y y etc. Example 4.. t u xe sin y Find the second partial derivatives uxy, uyx, u xx and the third partial derivative u ttt. t u e sin y x t u x e cos y y t uxy ux e cos y y t cos u u e y u x yx y xy t u x e sin y u t tt u u u u u u ttt Therefore u x e sin y ttt t t t u 0 xx
ENGI 344 4. Higher Partial Derivatives Page 4-08 Clairaut s Theorem If, on a disk D containing the point (a, b), f is defined and both of the partial derivatives f xy and f yx are continuous, (which is the case for most functions of interest), then xy,, f a b f a b that is, the order of differentiation doesn t matter. One of the most important partial differential equations involving second partial derivatives is Laplace s equation, which arises naturally in many applications, including electrostatics, fluid flow and heat conduction: u u 0 x y or its equivalent in 3 : u x u y yx u 0 Example 4.. Does u x y ln satisfy Laplace s equation? ln / ln n u x y x y x 0 x ux x y x y using ln x n ln x using x d f ln f x dx f x x y x x y x x y x y 0 u xx [using the quotient rule] u is symmetric with respect to x, y. Therefore, to find u yy, interchange x, y in u xx. x y uy y u y x u u 0 xx xx yy Therefore u does satisfy Laplace s equation (except at (0, 0)).
ENGI 344 4.3 Differentials Page 4-09 4.3 Differentials; error estimation; chain rule [again]; implicit functions; partial derivatives on curves of intersection In, let a curve have the Cartesian equation y = f (x). The small change in y, ( y ), caused by travelling along the curve for a small horiontal distance x, may be approximated by the change dy that is caused by travelling for the same horiontal distance x along the tangent line instead. The exact form is y f x x f x. The approximation to y is y dy f x dx where the increment x has been replaced by the differential dx. The approximation improves as x decreases towards ero. Stepping up one dimension, let a surface have the Cartesian equation = f (x, y). The change in the dependent variable caused by small changes in the independent variables x and y has the exact value f x x, y y f x, y. The approximation to is f f d dx dy x y Example 4.3. A rectangle has quoted dimensions of 30 cm for length and 4 cm for width. However, there may be an error of up to mm in the measurement of each dimension. Estimate the maximum error in the calculated area of the rectangle. Let A = area, L = length and W = width. Length = (30 ± 0.) cm L = 30 and L = 0. Width = (4 ± 0.) cm W = 4 and W = 0. A = LW A A da dl dw W dl L dw L W
ENGI 344 4.3 Differentials Page 4-0 Example 4.3. (continued) Let dl = dw = 0., then max(error) = A da = 4 0. + 30 0. = 5.4 cm. Compare this to a direct calculation: Amax A A Amin max error max, A A min 30 4 30 0. 4 0. = 5.39 cm. Amax A 30 0. 4 0. 30 4 = 5.4 cm. Therefore max(error) = 5.4 cm. Relative error: relative error in L = L 0. (one-third of one percent) L 30 300 relative error in W = W W 0. 4 40 A = LW da = W dl + L dw da W dl L dw A LW LW da dl dw 4 5 3 A L W 300 40 00 400 A da 0.75% A A and 0.75% of A = 70 cm is 5.4 cm.
ENGI 344 4.3 Differentials Page 4- Chain Rule f x, y. If x and y are both functions of t only, then, by the chain rule, d dx dy dt x dt y dt If f x, y and y in turn is a function of x only, then replace t by x in the formula above. dx dx and d dy dx x y dx Note the distinction between the total derivative d dx and the partial derivative x.
ENGI 344 4.3 Differentials Page 4- Example 4.3. In the study of fluid dynamics, one approach is to follow the motion of a point in the fluid. In that approach, the velocity vector is a function of both time and position, while, t r r t. position, in turn, is a function of time. v vr and The acceleration vector is then obtained through differentiation following the motion of the fluid: d v v v d x v d y v d a t dt t x dt y dt dt d v v or, equivalently, at v v dt t [The gradient operator,, will be introduced later, on page 4-.] Further analysis of an ideal fluid of density at pressure p subjected to a force field F leads to Euler s equation of motion v p v v F t This application of partial differentiation will be explored in some majors in a later x semester. As a simple example here, suppose that e t find the acceleration vector. v 0, then First note that the velocity vector is the derivative of the displacement vector, so that T v dr d x d y d x e 0 t dt dt dt dt T T v t T v x 0 0 0, e 0 0 x T v y v 0 0 0 T d v v v d x v d y v d a t dt t x dt y dt dt d v T T 0 0 0 x e 0 0 e x e x 0 0 dt 0 0 [One can show that xt ln t c, where c is a constant.] T
ENGI 344 4.3 Differentials Page 4-3 Generalied Chain Rule Let be a function of n variables x, x,, x n, each of which, in turn, is a function of m variables t, t,, t m, so that m m, x t, t,, t. f x t, t,, t, x t, t,, t, To find t i, n m trace all paths that start at and end at t i, via all of the n n j j jx j ti jx j ti x j variables. x d dx and i,,, m Example 4.3.3: st u x y y x, x st, y e and t. Find u s in terms of s and t only. Find the value of u s when s = 0 and t =. u u x u y s x s y s [There is no term because is not a function of s.] u st y 0 t x 0t e s st t e t st t e 3 t t st t e This derivative could also be found directly by replacing x, y and by the respective functions of s and t before differentiating u: st st u st e e t t st 3 3 st st st u st t e t e t t t t st t e s st st us e 0 0, 0 3.
ENGI 344 4.3 Differentials Page 4-4 Implicit functions: If is defined implicitly as a function of x and y by F x, y, df Fx dx Fy dy F d 0 F d Fx dx Fy dy provided F 0 F c, then Example 4.3.4: Find the change in when x and y both increase by 0. from the point (,, ) on the sphere x y 9. x dx y F x y 9 d dy x dx y dy Solution Approximate motion on the sphere by motion on the tangent plane: x =, y = =, dx Δx = 0., dy Δy = 0.. d 0. 0. 0.3 Therefore decreases by [approximately] 0.3 Exact: old 9 4, new 9...7.649 Therefore Δ = 0.3507...
ENGI 344 4.3 Differentials Page 4-5 Curves of Intersection Example 4.3.5: Find both partial derivatives with respect to on the curve of intersection of the sphere centre the origin, radius 5, and the circular cylinder, central axis on the y-axis, radius 3. Sphere: f = x + y + = 5 Cylinder: g = x + = 9 df = x dx + y dy + d = 0 and dg = x dx + d = 0 which leads to the linear system x y dx d x 0 dy dx x y 0 y d d dy x 0 xy x x y d d xy 0 x 0 Therefore dx d x and dy 0 d x y and x 0 [The intersection is the pair of circles x + = 9, y = 4. Because y never changes on each circle, x is actually a function of only.]
ENGI 344 4.3 Differentials Page 4-6 Example 4.3.6 f x, y, x y. Find A surface is defined by y. Implicit method: df f f f dx dy d x y 0 dx y dy x y d But f x y d dx y dy d dx y dy In the slice in which the partial derivative y is evaluated, x is constant dx = 0. 0 y y y Explicit method: x y x y 0 y 0 y y In general, if a dx b dy c d then a (because y is constant and dy = 0 in the slice in which x c x b (because x is constant and dx = 0 in the slice in which y c is evaluated) and is evaluated). y