13.4 TANGENT PLANES and DIFFERENTIALS

Transcription

1 3.4 Tangent Planes and Differentials Contemporar Calculus 3.4 TANGENT PLANES and DIFFERENTIALS In Section.8 we were able to use the derivative f ' of a function = f() of one variable to find the equation of the line tangent to the graph of f at a point (a, f(a)) (Fig. ): = f(a) + f '(a). ( a). And then we used this tangent line to approimate values of f near the point (a, f(a)), and we introduced the idea if the differential df = f '(a). d of the function f. In this section we etend these ideas to functions z = f(,) of two variables. But here we will find tangent planes (Fig. ) rather than tangent lines, and we will use the tangent plane to approimate values of f(,). Finall, we will etend the concept of a differential to functions of two variables. tangent line = f(a) + f '(a)( a) z = f(,) z tangent plane f(a) = f() a Fig. : Tangent line to = f() at = a a b (a,b) Fig. : Tangent plane to z = f(,) at (a,b) Tangent Planes In Section.6 we saw how to use a point ( a, b, c ) and two (nonparallel) vectors to determine the equation of the plane through the point and containing lines parallel to the given vectors (Fig. 3): () we used the cross product of the two given vectors to find a normal vector N = n, n, n 3 to the plane, and then N = U V (a,b,c) z Given the point (a,b,c) and vectors U and V V () we used the normal vector N and the point to write the equation of the plane as n ( a) + n ( b) + n 3 (z c) = 0. This approach also works when we need the equation of a plane tangent to a z = f(,), but we will use the formula for the to find the two needed vectors. U Equation of the tangent plane is n ( a) + n ( b) + n (z c) = 0 3 Fig. 3 0 z = f(,) = Eample : Find the equation of the plane tangent to the f(,) = at the point 0 P = (,, f(,) ) = (,, 9) on the (Fig. 4). Fig. 4 (,) 3

2 3.4 Tangent Planes and Differentials Contemporar Calculus Solution: We are given a point (,, 9) on the plane, but we need two vectors. These vectors are the rates of change of the f(,) in the and directions. The rate of change of f(,) in the direction is f (,) = 6, and at the point (,,9) we have f (,) = 6() = 6. Similarl, the rate of change of f(,) in the direction is f (,) =, and at the point (,,9) we have f (,) = () = 4. 0 Then a "rate of change vector in the direction" is U =, 0, 6 formed b taking "step" in the direction, taking 0 "steps" in the direction ( is constant), and then taking 6 "steps" in the z direction (6 = f (,) = rate of change of z with respect to increasing values). Similarl, a "rate of change vector in the direction is V = 0,, 4. These vectors are shown in Fig U V 3 (,) Fig. 5: Surface and tangent vectors U and V Now a normal vector N to the plane we want is formed b taking N = V U = i j k = (6)i ( 4)j + ( )k = 6i + 4j k (Note: taking N = U V = 6i 4j + k also works.) Finall, using the point P = (,, 9) and the normal vector N = V U = 6i + 4j k, we know that the equation of the plane is 6( ) + 4( ) (z 9) = 0 or z = 9 + 6( ) + 4( ) Looking at the equation z = 9 + 6( ) + 4( ) of the plane, ou should notice that the 9 is the z coordinate of our original point, that the coefficient of the variable, 6, is f (,), and that the coefficient of the variable, 4, is f (,). Fig. 6 shows the and the tangent plane. Fortunatel we do not need to go through all of those calculations ever time we need the equation of a plane tangent to a at a point: the pattern that we noted about the coefficients of the variables in the tangent plane equation and the values of the partial derivatives holds for ever differentiable function. 0 0 z = f(,) = (,) Fig. 6: Surface and tangent plane tangent plane z = 9 + 6( ) + 4( )

3 3.4 Tangent Planes and Differentials Contemporar Calculus 3 Equation for a Tangent Plane If f(,) is differentiable at the point (a, b, f(a,b) ), then the equation of the plane tangent to the z = f(, ) at the point P(a, b, f(a,b) ) is z = f(a,b) + f (a,b)( a) + f (a,b)( b). Proof: The proof simpl involves the steps we went through for Eample. U =, 0, f (a,b) is formed b taking "step" in the direction, taking 0 "steps" in the direction ( is constant), and then taking f (a,b) "steps" in the z direction. Similarl, V = 0,, f (a,b). Then N = V U = i j k 0 f (a,b) 0 f (a,b) = ( f (a,b) )i ( f (a,b) )j + ( )k = f (a,b) i + f (a,b) j k. Finall, using the point (a, b, f(a,b) ) and N = V U = f (a,b) i + f (a,b)j k, we have that the equation of the plane is f (a,b)( a) + f (a,b)( b) (z f(a,b)) = 0 or z = f(a,b) + f (a,b)( a) + f (a,b)( b). Eample : Find the plane tangent to the z = 3 + ln( ) + 7 at the point (,, 9). Solution: f (, ) = () = 43 + and f (, ) = 6 + () = 6 + so f (, ) = 5 and f (, ) = 7. Then the equation of the tangent plane is z = 9 + 5( ) + 7( ) or z = (This is much quicker than the "from scratch" method of Eample.) Fig. 7 shows two views of this and the tangent plane notice that in this case the tangent plane cuts through the z z = 3 + ln( ) z 0 tangent plane (,) z = 3 + ln( ) + 7 Fig. 7: Two views of the and tangent plane tangent plane Practice : Find the plane tangent to the z = sin( ) at the point (0,, 5).

4 3.4 Tangent Planes and Differentials Contemporar Calculus 4 Differentials The following "boed" material summarizes, from Section.8, the definition and results about the differential d of a function = f() of one variable. Definition for = f(): The differential of = f() is d = f '() d = df d d. Meaning of d: d is the change in the value, along the tangent line to f obtained b a step of d in the value. Result: If f is differentiable at = a and d is "small" then f(a + d) f(a) d or f(a + d) f(a) + d. Meaning of the Result: For a small step d, the actual change in f is approimatel equal to the change along the tangent line: f(a + d) f(a) + f '(a) d. The etension to functions of two variables is given in the net "bo." Definition for z = f(,): The differential (or total differential) of z = f(,) is dz = f (, ) d + f (, ) d = f d + f d. Meaning of dz: dz is the change in the z value, along the tangent plane to f, obtained b a step of d in the direction and a step of d in the direction. (Fig. 8) Result: If z = f(,) is differentiable at the point (a, b), then f(a + d, b + d) f(a,b) dz = f (a, b) d + f (a, b) d = f d + f d. Meaning of the Result: For a small step of d in the direction and a small step d in the direction, the change in f is approimatel equal to the change along the tangent plane: f(a + d, b + d) f(a,b) + dz = f(a,b) + f (a, b) d + f (a,b) d Eample 3: Find the differential of z = (a) in general, and (b) at the point (,) = (, ). Solution: (a) dz = f (a, b) d + f (a, b) d = { 6 3 } d + { 9 } d (b) at (, ), dz = { 6()() 3 } d + { 9() () } d = () d + (36) d. Eample 4: For z = and the point (,), use the result of Eample 3 that dz = () d + (36) d to approimate f(.0,.0 ) and f(.0, 0.97 ). Compare these approimate values with the eact values of f(.0,.0) and f(.0, 0.97).

5 3.4 Tangent Planes and Differentials Contemporar Calculus 5 Solution: For f(.0,.0 ), d = 0.0 and d = 0.0 so dz = ()(0.0) + (36)(0.0) = 0.6. Then f(.0,.0 ) f(, ) + dz = = 7.6. Actuall, f(.0,.0 ) = so the approimation "error" using the differential is 0.0. For f(.0, 0.97 ), d = 0.0 and d = 0.03, so dz = ()(0.0) + (36)( 0.03) = Then f(.0, 0.97 ) f(,) + dz = 7 + ( 0.96) = Actuall, f(.0, 0.97 ) = so the approimation "error" using the differential is 0.0. Practice : Find the differential of z = 3 +. sin( ) (a) in general, and (b) at the point (,) = (, π/). (c) Use the result of part (b) to approimate f(.3, π 0. ) and f( 0.99, π + 0.). (d) Compare the results of (c) with the eact values of f(.3, π 0. ) and f( 0.99, π + 0.). Eamples 4 and Practice (c and d) compare the value of f found b moving along the tangent plane to the actual value of f found on the. When the sidewas movement is "small" (when d and d are both small), then the "along the tangent plane" value of z is close to the "on the " value of z, the actual value of f. PROBLEMS In problems 8, find an equation for the tangent plane to the given at the given point.. z = + 4 at (,, 8). z = at (3,, 5) 3. z = 5 + ( ) + ( + ) at (, 0, 0) 4. z = sin( + ) at (,, 0) 5. z = ln( + ) at (, 3, 0) 6. z = e. ln( ) at (3,, 0) 7. z = at (,, ) 8. z = at (5,, ) In problems 9 8, find the differential of the given function. 9. z = 3 0. z = z = +. z =. e 3. u = e. cos( ) 4. v = ln( 3 ) 5. w = + z 6. w = sin( z ) 7. w = ln( + + z ) 8. w = + + z 9. If z = 5 + and (, ) changes from (, ) to (.05,.), compare the values of z and dz. 0. If z = + 3 and (, ) changes from (3, ) to (.96, 0.95), compare the values of z and dz.

6 3.4 Tangent Planes and Differentials Contemporar Calculus 6 In problems 4, use differentials to approimate the value of f at the given point.. f(, ) = at (5.0, 4.0). f(, ) = 0 7 at (.95,.08) 3. f(, ) = ln( 3 ) at (6.9,.06) 4. f(, ) =. e at (0.,.96) 5. The length and width of a rectangle are measured as 30 cm and 4 cm, respectivel, with an error in measurement of at most 0. cm in each. Use differentials to estimate the maimum error in the calculated area of the rectangle. 6. The dimensions of a closed rectangular bo are measured as 80 cm, 60 cm, and 50 cm, respectivel, with a possible error of 0. cm in each dimension. Use differentials to estimate the maimum error in calculating the area of the bo. 7. Use differentials to estimate the amount of tin in a closed tin can with diameter 8 cm and height cm if the tin is 0.04 cm thick. 8. Use differentials to estimate the amount of metal in a closed clindrical can that is 0 cm high and 4 cm in diameter if the metal in the wall is 0.05 cm thick and the metal in the top and bottom is 0. cm thick. 9. A boundar stripe 3 in. wide is painted around a rectangle whose dimensions are 00 ft. b 00 ft. Use differentials to approimate the number of square feet of paint in the stripe. 30. The pressure, volume, and temperature of a mole of an ideal gas are related b the equation PV = 8.3T, where P is measured in kilopascals, V in liters, and T in o K (= o C + 73). Use differentials to find the approimate change in pressure if the volume increases from L to.3 L and the temperature decreases from 30 o K to 305 o K. PRACTICE ANSWERS Practice : z = f(a,b) + f (a,b)( a) + f (a,b)( b) with f(,) = sin( ), a = 0, b =, and f(0,) = 5. f (,) = 5 +. cos( ) so f (0,) = 5 + = 6. f (,) = cos( ) so f (0,) = 7. Then the equation of the tangent plane to f at (0,,5) is z = 5 + 6( 0) + 7( ) =

7 3.4 Tangent Planes and Differentials Contemporar Calculus 7 Practice : (a) dz = f (a, b) d + f (a, b) d = {. cos( ). + sin( )} d + {. cos( ). } d (b) at (, π/), dz = {. cos(.. π/ ).. π/ + sin(.. π/ )} d + {. cos(.. π/ ).. } d so dz = ( π) d + ( ) d. (c)&(d) For f(.3, π 0. ), d = 0.3 and d = 0. so dz = ( π)( 0.3 ) + ( )( 0.) Then f(.3, π 0. ) f(, π ) + dz = 3 + ( 0.74) =.58. Actuall, f(.3, π 0. ) = so the approimation "error" is 0.078, an "error" of less than 4%. For f( 0.99, π + 0.), d = 0.0 and d = 0. so dz = ( π)( 0.0) + ( )(0.) = Then f( 0.99, π + 0.) f(, π ) + dz = 3 + ( 0.369) =.63. Actuall, f( 0.99, π + 0.) = so the approimation "error" is 0.06, an "error" of less than %. Selected Answers z = z + 6 = z = 7. z + = 0 9. dz = 3 d + 3 d. dz = ( + ) d ( + ) d 3. du = e. ( cos( ). sin( ) ) d. e. sin( ) d 5. dw = d + ( + z) d + dz 7. dw = ( + + z ) ( d + d + z dz) 9. z = 0.95 and dz = cm 7. 6 cm

8 3.4 Tangent Planes and Differentials Contemporar Calculus 8 Appendi: Maple commands to create figures In the following Maple routines: a and a are the and coordinates, respectivel, of the point of tangenc, the formula for za is our function (in terms of a and a), f is f (in terms of a and a), and f is f (in terms of a and a). L is the length and width of one of the rectangles making up the plane, and N is the number of rectangles in the plane to each side of the point. min, ma, min, ma, zmin, and zma specif the viewing "rectange in 3D. orientation= [θ degrees, φ degrees] is the viewing position in spherical coordinates. > with(plots); > MAKES FIG. a:=.6: a:=:za:=6-*a^-a^: f:=-4*a: f:=-*a: L:=0.3:N:=: min:=0: ma:=.5: min:=0: ma:=3: zmin:=0: zma:=6: SURF:=plot3d(6-*^-^, =min..ma, =min..ma, aes=normal, grid=[9,7], color=red): PT:=pointplot3d( {[a,a,za]}, color=black, smbol=circle): TANPL:=plot3d( [u,v,za+f*(u-a)+f*(v-a)], u=a-n*l..a+n*l, v=a-n*l..a+n*l, color=blue, grid=[*n+,*n+], thickness=): displa3d( {SURF, PT, TANPL }, orientation=[5,80], tickmarks=[,,4], view=0..6); MAKES FIG. 4 (these commands also do all of the work for figures 5 and 6) To modif these commands for our own function at a point ou pick, ou onl need to change the part of the routine in boldface tpe. > a:=: a:=:za:=*a^3+a^+3: f:=6*a^: f:=*a: L:=0.3:N:=: min:=0: ma:=: min:=0: ma:=3: zmin:=0: zma:=5: SURF:=plot3d(*^3+^+3, =min..ma, =min..ma, aes=normal, grid=[9,7], color=red): PT:=pointplot3d( {[a,a,za]}, color=black, smbol=circle): TANPL:=plot3d( [u,v,za+f*(u-a)+f*(v-a)], u=a-n*l..a+n*l, v=a-n*l..a+n*l, color=blue, grid=[*n+,*n+], thickness=): DX:=spacecurve( [u,a,za+f*(u-a)],u=a..a+3*l, color=black, thickness=): DY:=spacecurve( [a,v,za+f*(v-a)],v=a..a+3*l, color=black, thickness=): displa3d( {SURF, PT }, orientation=[5,80], tickmarks=[3,3,4], view=0..zma); > MAKES FIG. 5 displa3d( {SURF, PT, DX, DY }, orientation=[5,80], tickmarks=[3,3,4], view=0..zma); > MAKES FIG. 6 two views displa3d( {SURF, PT, DX, DY, TANPL }, orientation=[5,80], tickmarks=[3,3,4], view=0..zma); displa3d( {SURF, PT, DX, DY, TANPL }, orientation=[70,57], tickmarks=[3,3,4], view=0..zma); MAKES FIG. 7 two views > a:=: a:=:za:=*a^*a^3+ln(a*a)+7: f:=4*a*a^3 +/a: f:=6*a^*a^+/a: L:=0.3:N:=: min:=.: ma:=: min:=.: ma:=: zmin:=0: zma:=30: SURF:=plot3d(*^*^3+ln(*)+7, =min..ma, =min..ma, aes=normal, grid=[9,7], color=red): PT:=pointplot3d( {[a,a,za]}, color=black, smbol=circle): TANPL:=plot3d( [u,v,za+f*(u-a)+f*(v-a)], u=a-n*l..a+n*l, v=a-n*l..a+n*l, color=blue, grid=[*n+,*n+], thickness=): displa3d( {SURF, PT, TANPL}, orientation=[30,80], tickmarks=[3,3,4], view=0..zma); displa3d( {SURF, PT, TANPL}, orientation=[0,5], tickmarks=[3,3,4], view=0..zma);