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# 17 tangent planes and total differentials

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### 17 tangent planes and total differentials

1. 1. Tangent Planes
2. 2. Tangent Planes zRecall that a surface z = f(x, y) is P =(a, b, c)differentiable at a point P = (a, b, c)or differentiable at p = (a, b), yif it is smooth (no crease) and has a x (a, b)well defined tangent plane at P. A smooth point P and its tangent plane.
3. 3. Tangent Planes zRecall that a surface z = f(x, y) is P =(a, b, c)differentiable at a point P = (a, b, c)or differentiable at p = (a, b), yif it is smooth (no crease) and has a x (a, b)well defined tangent plane at P. A smooth point P and its tangent plane.An "elementary function" is differentiable at a point pif there is a circle centered at p and that all the partialsexist in this circle.
4. 4. Tangent Planes zRecall that a surface z = f(x, y) is P =(a, b, c)differentiable at a point P = (a, b, c)or differentiable at p = (a, b), yif it is smooth (no crease) and has a x (a, b)well defined tangent plane at P. A smooth point P and its tangent plane.An "elementary function" is differentiable at a point pif there is a circle centered at p and that all the partialsexist in this circle.Just as the tangent line to a curve at a pointapproximates the curve around the point, the tangentplane to a surface at a point P approximates thesurface around P.
5. 5. Tangent Planes zRecall that a surface z = f(x, y) is P =(a, b, c)differentiable at a point P = (a, b, c)or differentiable at p = (a, b), yif it is smooth (no crease) and has a x (a, b)well defined tangent plane at P. A smooth point P and its tangent plane.An "elementary function" is differentiable at a point pif there is a circle centered at p and that all the partialsexist in this circle.Just as the tangent line to a curve at a pointapproximates the curve around the point, the tangentplane to a surface at a point P approximates thesurface around P.To find an equation for the tangent plane at P, we finda normal vector N for the plane in the following
6. 6. Tangent Planes zWe note that a vector that sits in the <x, 0, z>plane y = b has the form <x, 0, z>. y=b x y
7. 7. Tangent Planes zWe note that a vector that sits in the <x, 0, z>plane y = b has the form <x, 0, z>. y=bSimilarly, a vector that sits in the x yplane x = a has the form < 0, y, z>.
8. 8. Tangent PlanesWe note that a vector that sits in the <x, 0, z> zplane y = b has the form <x, 0, z>. y=bSimilarly, a vector that sits in theplane x = a has the form < 0, y, z>. x yThe tangent line to the y–trace at P = (a, b, c)
9. 9. Tangent PlanesWe note that a vector that sits in the <x, 0, z> zplane y = b has the form <x, 0, z>. y=bSimilarly, a vector that sits in theplane x = a has the form < 0, y, z>. x yThe tangent line to the y–trace at P = (a, b, c) z (a, b, c) y y=b x
10. 10. Tangent PlanesWe note that a vector that sits in the <x, 0, z> zplane y = b has the form <x, 0, z>. y=bSimilarly, a vector that sits in theplane x = a has the form < 0, y, z>. x yThe tangent line to the y–trace at P = (a, b, c) z (a, b, c) y y=b x
11. 11. Tangent PlanesWe note that a vector that sits in the <x, 0, z> zplane y = b has the form <x, 0, z>. y=bSimilarly, a vector that sits in theplane x = a has the form < 0, y, z>. x yThe tangent line to the y–trace at P = (a, b, c) is in thetangent plane T and it has slope dz/dx|p = M. z (a, b, c) y y=b x
12. 12. Tangent PlanesWe note that a vector that sits in the <x, 0, z> zplane y = b has the form <x, 0, z>. y=bSimilarly, a vector that sits in theplane x = a has the form < 0, y, z>. x yThe tangent line to the y–trace at P = (a, b, c) is in thetangent plane T and it has slope dz/dx|p = M. z T (a, b, c) y y=b x
13. 13. Tangent PlanesWe note that a vector that sits in the <x, 0, z> zplane y = b has the form <x, 0, z>. y=bSimilarly, a vector that sits in theplane x = a has the form < 0, y, z>. x yThe tangent line to the y–trace at P = (a, b, c) is in thetangent plane T and it has slope dz/dx|p = M. z T z (a, c) y=b (a, b, c) z = f(x, b) x y y=b x
14. 14. Tangent PlanesWe note that a vector that sits in the <x, 0, z> zplane y = b has the form <x, 0, z>. y=bSimilarly, a vector that sits in theplane x = a has the form < 0, y, z>. x yThe tangent line to the y–trace at P = (a, b, c) is in thetangent plane T and it has slope dz/dx|p = M. z T z (a, c) y=b 1 (a, b, c) z = f(x, b) M =d z d x x y y=b x
15. 15. Tangent PlanesWe note that a vector that sits in the <x, 0, z> zplane y = b has the form <x, 0, z>. y=bSimilarly, a vector that sits in theplane x = a has the form < 0, y, z>. x yThe tangent line to the y–trace at P = (a, b, c) is in thetangent plane T and it has slope dz/dx|p = M.Hence the vector <1, 0, M> is in the tangent plane T. z T z (a, c) y=b 1 (a, b, c) z = f(x, b) M =d z d x x y y=b x
16. 16. Tangent PlanesSimilarly, the tangent plane must contain the tangentline in the y-direction whose slope is dz/dy|p= L . z x=a x y
17. 17. Tangent PlanesSimilarly, the tangent plane must contain the tangentline in the y-direction whose slope is dz/dy|p= L . z T (a, b, c) x=a x y
18. 18. Tangent PlanesSimilarly, the tangent plane must contain the tangentline in the y-direction whose slope is dz/dy|p= L . z T z x=a (a, b, c) z = f(a,y) 1 (b, c) L x=a y x y
19. 19. Tangent PlanesSimilarly, the tangent plane must contain the tangentline in the y-direction whose slope is dz/dy|p= L .Hence the vector <0, 1, L> is a tangent vector for thex trace and <0, 1, L> is in T. z T z x=a (a, b, c) z = f(a,y) 1 (b, c) L x=a y x y
20. 20. Tangent PlanesSimilarly, the tangent plane must contain the tangentline in the y-direction whose slope is dz/dy|p= L .Hence the vector <0, 1, L> is a tangent vector for thex trace and <0, 1, L> is in T. z T z x=a (a, b, c) z = f(a,y) 1 (b, c) L x=a y x y zSince <1, 0, M > and <0, 1, L>are in T so their cross product P<1, 0, M> x <0, 1, L> = <–M, –L, 1>is a normal vector to the tangent <1, 0, M> <0, 1, L>plane T. x y
21. 21. Tangent PlanesSimilarly, the tangent plane must contain the tangentline in the y-direction whose slope is dz/dy|p= L .Hence the vector <0, 1, L> is a tangent vector for thex trace and <0, 1, L> is in T. z T z x=a (a, b, c) z = f(a,y) 1 (b, c) L x=a y x y zSince <1, 0, M > and <0, 1, L> <–M, –L, 1>are in T so their cross product P<1, 0, M> x <0, 1, L> = <–M, –L, 1>is a normal vector to the tangent <1, 0, M> <0, 1, L>plane T. x y
22. 22. Tangent PlanesEquation of The Tangent Plane. Given that a normalvector is <–M, –L, 1>, an equation of the tangent planeat P = (a, b, c) is (z – c) = M(x – a) + L(y – b)where with M = dz / dx|p and L = dz / dy|p <–M, –L, 1> P <0, 1, L> <1, 0, M> x y
23. 23. Tangent PlanesEquation of The Tangent Plane. Given that a normalvector is <–M, –L, 1>, an equation of the tangent planeat P = (a, b, c) is (z – c) = M(x – a) + L(y – b)where with M = dz / dx|p and L = dz / dy|p <–M, –L, 1>Example A. Find the equation of the Ptangent plane for z = √49 – x2 – y2 at <0, 1, L> <1, 0, M>(2, 3, 6). x y
24. 24. Tangent PlanesEquation of The Tangent Plane. Given that a normalvector is <–M, –L, 1>, an equation of the tangent planeat P = (a, b, c) is (z – c) = M(x – a) + L(y – b)where with M = dz / dx|p and L = dz / dy|p <–M, –L, 1>Example A. Find the equation of the Ptangent plane for z = √49 – x2 – y2 at <0, 1, L> <1, 0, M>(2, 3, 6). x yfx(P)= –x/√49 – x2 – y2, fy(P)= –x/√49 – x2 – y2,
25. 25. Tangent PlanesEquation of The Tangent Plane. Given that a normalvector is <–M, –L, 1>, an equation of the tangent planeat P = (a, b, c) is (z – c) = M(x – a) + L(y – b)where with M = dz / dx|p and L = dz / dy|p <–M, –L, 1>Example A. Find the equation of the Ptangent plane for z = √49 – x2 – y2 at <0, 1, L> <1, 0, M>(2, 3, 6). x yfx(P)= –x/√49 – x2 – y2, fy(P)= –x/√49 – x2 – y2,Hence M = fx(2, 3) = –1/3, and L = fy(2, 3) = –1/2
26. 26. Tangent PlanesEquation of The Tangent Plane. Given that a normalvector is <–M, –L, 1>, an equation of the tangent planeat P = (a, b, c) is (z – c) = M(x – a) + L(y – b)where with M = dz / dx|p and L = dz / dy|p <–M, –L, 1>Example A. Find the equation of the Ptangent plane for z = √49 – x2 – y2 at <0, 1, L> <1, 0, M>(2, 3, 6). x yfx(P)= –x/√49 – x2 – y2, fy(P)= –x/√49 – x2 – y2,Hence M = fx(2, 3) = –1/3, and L = fy(2, 3) = –1/2So the equation of the tangent plane at (2,3,6) is(z – 6) = –1/3(x – 2) – 1/2 (y – 3)or that 2x + 3y + 6z = 49.
27. 27. Tangent PlanesEquation of The Tangent Plane. Given that a normalvector is <–M, –L, 1>, an equation of the tangent planeat P = (a, b, c) is (z – c) = M(x – a) + L(y – b)where with M = dz / dx|p and L = dz / dy|p <–M, –L, 1>Example A. Find the equation of the Ptangent plane for z = √49 – x2 – y2 at <0, 1, L> <1, 0, M>(2, 3, 6). x yfx(P)= –x/√49 – x2 – y2, fy(P)= –x/√49 – x2 – y2,Hence M = fx(2, 3) = –1/3, and L = fy(2, 3) = –1/2So the equation of the tangent plane at (2,3,6) is(z – 6) = –1/3(x – 2) – 1/2 (y – 3)or that 2x + 3y + 6z = 49.Use a graphing software to verity this result.
28. 28. Total DifferentialsLet y = f(x), we write Δx as dx to represent a smallchange in the value x, and we write Δy = f(x + Δx) – f(x)as the corresponding change in the output y.
29. 29. Total DifferentialsLet y = f(x), we write Δx as dx to represent a smallchange in the value x, and we write Δy = f(x + Δx) – f(x)as the corresponding change in the output y.If f(x) is differentiable at x = a, thenf(a + dx) ≈ f(a) + f(a)*dx where f(a)*dx ≈ Δy.
30. 30. Total DifferentialsLet y = f(x), we write Δx as dx to represent a smallchange in the value x, and we write Δy = f(x + Δx) – f(x)as the corresponding change in the output y.If f(x) is differentiable at x = a, thenf(a + dx) ≈ f(a) + f(a)*dx where f(a)*dx ≈ Δy. y=f(x) f(a) a
31. 31. Total DifferentialsLet y = f(x), we write Δx as dx to represent a smallchange in the value x, and we write Δy = f(x + Δx) – f(x)as the corresponding change in the output y.If f(x) is differentiable at x = a, thenf(a + dx) ≈ f(a) + f(a)*dx where f(a)*dx ≈ Δy. y=f(x) f(a) f(a+dx) dx a a+dx
32. 32. Total DifferentialsLet y = f(x), we write Δx as dx to represent a smallchange in the value x, and we write Δy = f(x + Δx) – f(x)as the corresponding change in the output y.If f(x) is differentiable at x = a, thenf(a + dx) ≈ f(a) + f(a)*dx where f(a)*dx ≈ Δy. Δy y=f(x) f(a) f(a+dx) dx a a+dx
33. 33. Total DifferentialsLet y = f(x), we write Δx as dx to represent a smallchange in the value x, and we write Δy = f(x + Δx) – f(x)as the corresponding change in the output y.If f(x) is differentiable at x = a, thenf(a + dx) ≈ f(a) + f(a)*dx where f(a)*dx ≈ Δy. Δy y=f(x) f(a) f(a+dx) dx a a+dx
34. 34. Total DifferentialsLet y = f(x), we write Δx as dx to represent a smallchange in the value x, and we write Δy = f(x + Δx) – f(x)as the corresponding change in the output y.If f(x) is differentiable at x = a, thenf(a + dx) ≈ f(a) + f(a)*dx where f(a)*dx ≈ Δy. f(a) dx=dy * Δy y=f(x) f(a) f(a+dx) dx a a+dx
35. 35. Total DifferentialsLet y = f(x), we write Δx as dx to represent a smallchange in the value x, and we write Δy = f(x + Δx) – f(x)as the corresponding change in the output y.If f(x) is differentiable at x = a, thenf(a + dx) ≈ f(a) + f(a)*dx where f(a)*dx ≈ Δy. f(a) dx=dy * Δy y=f(x) f(a) f(a+dx) f(a)+ f(a)*dx dx a a+dx
36. 36. Total DifferentialsLet y = f(x), we write Δx as dx to represent a smallchange in the value x, and we write Δy = f(x + Δx) – f(x)as the corresponding change in the output y.If f(x) is differentiable at x = a, thenf(a + dx) ≈ f(a) + f(a)*dx where f(a)*dx ≈ Δy. f(a) dx=dy *We call dx “the differential of x”, Δy y=f(x)and we write the approximationf(x)dx as dy which is called f(a) f(a+dx) f(a)+ f(a)*dx“the differential of y”. We have that dxf(a)*dx = dy ≈ Δy. a a+dx
37. 37. Total DifferentialsLet y = f(x), we write Δx as dx to represent a smallchange in the value x, and we write Δy = f(x + Δx) – f(x)as the corresponding change in the output y.If f(x) is differentiable at x = a, thenf(a + dx) ≈ f(a) + f(a)*dx where f(a)*dx ≈ Δy. f(a) dx=dy *We call dx “the differential of x”, Δy y=f(x)and we write the approximationf(x)dx as dy which is called f(a) f(a+dx) f(a)+ f(a)*dx“the differential of y”. We have that dxf(a)*dx = dy ≈ Δy. a a+dxWe use dy, which is easier to calculate, to estimatesthe change in the output.
38. 38. Total DifferentialsLet y = f(x), we write Δx as dx to represent a smallchange in the value x, and we write Δy = f(x + Δx) – f(x)as the corresponding change in the output y.If f(x) is differentiable at x = a, thenf(a + dx) ≈ f(a) + f(a)*dx where f(a)*dx ≈ Δy. f(a) dx=dy *We call dx “the differential of x”, Δy y=f(x)and we write the approximationf(x)dx as dy which is called f(a) f(a+dx) f(a)+ f(a)*dx“the differential of y”. We have that dxf(a)*dx = dy ≈ Δy. a a+dxWe use dy, which is easier to calculate, to estimatesthe change in the output. For example, let y = f(x) = x 2,at (3, 9), dy = f(3) dx = 6 dx,.
39. 39. Total DifferentialsLet y = f(x), we write Δx as dx to represent a smallchange in the value x, and we write Δy = f(x + Δx) – f(x)as the corresponding change in the output y.If f(x) is differentiable at x = a, thenf(a + dx) ≈ f(a) + f(a)*dx where f(a)*dx ≈ Δy. f(a) dx=dy *We call dx “the differential of x”, Δy y=f(x)and we write the approximationf(x)dx as dy which is called f(a) f(a+dx) f(a)+ f(a)*dx“the differential of y”. We have that dxf(a)*dx = dy ≈ Δy. a a+dxWe use dy, which is easier to calculate, to estimatesthe change in the output. For example, let y = f(x) = x 2,at (3, 9), dy = f(3) dx = 6 dx. So if dx = 0.01,then dy = 0.06.
40. 40. Total DifferentialsLet y = f(x), we write Δx as dx to represent a smallchange in the value x, and we write Δy = f(x + Δx) – f(x)as the corresponding change in the output y.If f(x) is differentiable at x = a, thenf(a + dx) ≈ f(a) + f(a)*dx where f(a)*dx ≈ Δy. f(a) dx=dy *We call dx “the differential of x”, Δy y=f(x)and we write the approximationf(x)dx as dy which is called f(a) f(a+dx) f(a)+ f(a)*dx“the differential of y”. We have that dxf(a)*dx = dy ≈ Δy. a a+dxWe use dy, which is easier to calculate, to estimatesthe change in the output. For example, let y = f(x) = x 2,at (3, 9), dy = f(3) dx = 6 dx. So if dx = 0.01,then dy = 0.06. Note that Δy = f(3.01) – f(3) = 0.0601.
41. 41. Total DifferentialsLet (a, b, c) be a point in the plane z = Mx + Ly + k. z y (a, b ,0) x
42. 42. Total DifferentialsLet (a, b, c) be a point in the plane z = Mx + Ly + k.Let dx be a small increment in the x coordinate,and dy be a small increment in the y coordinate. z (a+dx, dy y b+dy,0) dx (a, b ,0) (a+dx, b ,0) x
43. 43. Total DifferentialsLet (a, b, c) be a point in the plane z = Mx + Ly + k.Let dx be a small increment in the x coordinate,and dy be a small increment in the y coordinate.Then the corresponding point in the plane is(a+dx, b+dy, c+dz) where dz = M*dx+L*dy (a+dx, b+dy, c+dz) z (a+dx, dy y b+dy,0) dx (a, b ,0) (a+dx, b ,0) x
44. 44. Total DifferentialsLet (a, b, c) be a point in the plane z = Mx + Ly + k.Let dx be a small increment in the x coordinate,and dy be a small increment in the y coordinate.Then the corresponding point in the plane is(a+dx, b+dy, c+dz) where dz = M*dx+L*dy with M = slope in the x direction, (a+dx, b+dy, c+dz) M*dx z (a+dx, dy y b+dy,0) dx (a, b ,0) (a+dx, b ,0) x
45. 45. Total DifferentialsLet (a, b, c) be a point in the plane z = Mx + Ly + k.Let dx be a small increment in the x coordinate,and dy be a small increment in the y coordinate.Then the corresponding point in the plane is(a+dx, b+dy, c+dz) where dz = M*dx+L*dy with M = slope in the x direction, and (a+dx, b+dy, c+dz) L = slope in the y direction. L*dy M*dx M*dx z (a+dx, dy y b+dy,0) dx (a, b ,0) (a+dx, b ,0) x
46. 46. Total DifferentialsLet (a, b, c) be a point in the plane z = Mx + Ly + k.Let dx be a small increment in the x coordinate,and dy be a small increment in the y coordinate.Then the corresponding point in the plane is(a+dx, b+dy, c+dz) where dz = M*dx+L*dy with M = slope in the x direction, and (a+dx, b+dy, c+dz) L = slope in the y direction. L*dy dz M*dx M*dx z (a+dx, dy y b+dy,0) dx (a, b ,0) (a+dx, b ,0) x
47. 47. Total DifferentialsLet (a, b, c) be a point in the plane z = Mx + Ly + k.Let dx be a small increment in the x coordinate,and dy be a small increment in the y coordinate.Then the corresponding point in the plane is(a+dx, b+dy, c+dz) where dz = M*dx+L*dy with M = slope in the x direction, and (a+dx, b+dy, c+dz) L = slope in the y direction. Geometrically, this says L*dy that if we crawl along the exterior walls of a M*dx building, the total height M*dx gained is the sum of z dy (a+dx, y b+dy,0) the heights gained (a, b ,0) dx (a+dx, b ,0) along each face. x
48. 48. Total DifferentialsGiven z = f(x, y) and that it’s differentiable atP = (a, b, c) and T be the tangent plane at P. z P=(a, b, c) z = f(x, y) (a, b) y x
49. 49. Total DifferentialsGiven z = f(x, y) and that it’s differentiable atP = (a, b, c) and T be the tangent plane at P.The point q = (a+dx, b+dy, c+dz) in T approximate the location Q = (a+dx, b+dy, c+Δz) on the surface ofz = f(x, y). z P=(a, b, c) z = f(x, y) (a, b) y x
50. 50. Total DifferentialsGiven z = f(x, y) and that it’s differentiable atP = (a, b, c) and T be the tangent plane at P.The point q = (a+dx, b+dy, c+dz) in T approximate the location Q = (a+dx, b+dy, c+Δz) on the surface ofz = f(x, y). z Q=(a+dx, b+dy, c+Δz) P=(a, b, c) z = f(x, y) (a, b) y x dx dy (a+dx, b+dy)
51. 51. Total DifferentialsGiven z = f(x, y) and that it’s differentiable atP = (a, b, c) and T be the tangent plane at P.The point q = (a+dx, b+dy, c+dz) in T approximate the location Q = (a+dx, b+dy, c+Δz) on the surface ofz = f(x, y). z q =(a+dx, b+dy, c+dz) Q=(a+dx, b+dy, c+Δz) P=(a, b, c) z = f(x, y) (a, b) y x dx dy (a+dx, b+dy)
52. 52. Total DifferentialsGiven z = f(x, y) and that it’s differentiable atP = (a, b, c) and T be the tangent plane at P.The point q = (a+dx, b+dy, c+dz) in T approximate the location Q = (a+dx, b+dy, c+Δz) on the surface ofz = f(x, y). The difference in the z q =(a+dx, b+dy, c+dz)z–value between P and q isdz = M*dx + L* dy where P=(a, b, c) Q=(a+dx, b+dy, c+Δz)M = df /dx|(a,b) z = f(x, y)= the slope in the x direction, (a, b) y x dxL = df / dx|(a,b) dy (a+dx, b+dy)= the slope in the y direction. q =(a+dx, b+dy, c+dz) Q=(a+dx, b+dy, c+Δz) P=(a, b, c) Close up
53. 53. Total DifferentialsGiven z = f(x, y) and that it’s differentiable atP = (a, b, c) and T be the tangent plane at P.The point q = (a+dx, b+dy, c+dz) in T approximate the location Q = (a+dx, b+dy, c+Δz) on the surface ofz = f(x, y). The difference in the z q =(a+dx, b+dy, c+dz)z–value between P and q isdz = M*dx + L* dy where P=(a, b, c) Q=(a+dx, b+dy, c+Δz)M = df /dx|(a,b) z = f(x, y)= the slope in the x direction, (a, b) y x dxL = df / dx|(a,b) dy (a+dx, b+dy)= the slope in the y direction. q =(a+dx, b+dy, c+dz) Q=(a+dx, b+dy, c+Δz) dz = M*dx + L* dy P=(a, b, c) Close up
54. 54. Total DifferentialsGiven z = f(x, y) and that it’s differentiable atP = (a, b, c) and T be the tangent plane at P.The point q = (a+dx, b+dy, c+dz) in T approximate the location Q = (a+dx, b+dy, c+Δz) on the surface ofz = f(x, y). The difference in the z q =(a+dx, b+dy, c+dz)z–value between P and q isdz = M*dx + L* dy where P=(a, b, c) Q=(a+dx, b+dy, c+Δz)M = df /dx|(a,b) z = f(x, y)= the slope in the x direction, (a, b) y x dxL = df / dx|(a,b) dy (a+dx, b+dy)= the slope in the y direction. q =(a+dx, b+dy, c+dz)The quantity dz = M*dx + L* dy Q=(a+dx, b+dy, c+Δz)is called the total differential dz = M*dx + L* dy P=(a, b, c)of f(x, y). Close up
55. 55. Total DifferentialsExample B. Approximate the value of x3y – x2y2at (x, y) = (1.03, 2.94) using the differentials.
56. 56. Total DifferentialsExample B. Approximate the value of x3y – x2y2at (x, y) = (1.03, 2.94) using the differentials.Let’s set z = f(x,y) = x3y – x2y2.
57. 57. Total DifferentialsExample B. Approximate the value of x3y – x2y2at (x, y) = (1.03, 2.94) using the differentials.Let’s set z = f(x,y) = x3y – x2y2. Select P = (1, 3)which is close to (1.03, 2.94) and easy to evaluate.
58. 58. Total DifferentialsExample B. Approximate the value of x3y – x2y2at (x, y) = (1.03, 2.94) using the differentials.Let’s set z = f(x,y) = x3y – x2y2. Select P = (1, 3)which is close to (1.03, 2.94) and easy to evaluate.At P = (1, 3), z = 3 – 9 = –6.
59. 59. Total DifferentialsExample B. Approximate the value of x3y – x2y2at (x, y) = (1.03, 2.94) using the differentials.Let’s set z = f(x,y) = x3y – x2y2. Select P = (1, 3)which is close to (1.03, 2.94) and easy to evaluate.At P = (1, 3), z = 3 – 9 = –6. To find an approximatevalue of f(1.03, 2.94), we calculate the z–value of(1.03, 2.94) on the tangent plane that touches (1, 3, 6).
60. 60. Total DifferentialsExample B. Approximate the value of x3y – x2y2at (x, y) = (1.03, 2.94) using the differentials.Let’s set z = f(x,y) = x3y – x2y2. Select P = (1, 3)which is close to (1.03, 2.94) and easy to evaluate.At P = (1, 3), z = 3 – 9 = –6. To find an approximatevalue of f(1.03, 2.94), we calculate the z–value of(1.03, 2.94) on the tangent plane that touches (1, 3, 6).From (1, 3) to (1.03, 2.94), dx = 0.03 and dy = –0.06.
61. 61. Total DifferentialsExample B. Approximate the value of x3y – x2y2at (x, y) = (1.03, 2.94) using the differentials.Let’s set z = f(x,y) = x3y – x2y2. Select P = (1, 3)which is close to (1.03, 2.94) and easy to evaluate.At P = (1, 3), z = 3 – 9 = –6. To find an approximatevalue of f(1.03, 2.94), we calculate the z–value of(1.03, 2.94) on the tangent plane that touches (1, 3, 6).From (1, 3) to (1.03, 2.94), dx = 0.03 and dy = –0.06.The partials at (1, 3), M = fx|(1,3)= 3x2y – 2xy2|(1,3) = –9,
62. 62. Total DifferentialsExample B. Approximate the value of x3y – x2y2at (x, y) = (1.03, 2.94) using the differentials.Let’s set z = f(x,y) = x3y – x2y2. Select P = (1, 3)which is close to (1.03, 2.94) and easy to evaluate.At P = (1, 3), z = 3 – 9 = –6. To find an approximatevalue of f(1.03, 2.94), we calculate the z–value of(1.03, 2.94) on the tangent plane that touches (1, 3, 6).From (1, 3) to (1.03, 2.94), dx = 0.03 and dy = –0.06.The partials at (1, 3), M = fx|(1,3)= 3x2y – 2xy2|(1,3) = –9,and L = fy|(1,3)= x3 – 2x2y|(1,3) = –5.
63. 63. Total DifferentialsExample B. Approximate the value of x3y – x2y2at (x, y) = (1.03, 2.94) using the differentials.Let’s set z = f(x,y) = x3y – x2y2. Select P = (1, 3)which is close to (1.03, 2.94) and easy to evaluate.At P = (1, 3), z = 3 – 9 = –6. To find an approximatevalue of f(1.03, 2.94), we calculate the z–value of(1.03, 2.94) on the tangent plane that touches (1, 3, 6).From (1, 3) to (1.03, 2.94), dx = 0.03 and dy = –0.06.The partials at (1, 3), M = fx|(1,3)= 3x2y – 2xy2|(1,3) = –9,and L = fy|(1,3)= x3 – 2x2y|(1,3) = –5.Hence dz = M*dx + L* dy
64. 64. Total DifferentialsExample B. Approximate the value of x3y – x2y2at (x, y) = (1.03, 2.94) using the differentials.Let’s set z = f(x,y) = x3y – x2y2. Select P = (1, 3)which is close to (1.03, 2.94) and easy to evaluate.At P = (1, 3), z = 3 – 9 = –6. To find an approximatevalue of f(1.03, 2.94), we calculate the z–value of(1.03, 2.94) on the tangent plane that touches (1, 3, 6).From (1, 3) to (1.03, 2.94), dx = 0.03 and dy = –0.06.The partials at (1, 3), M = fx|(1,3)= 3x2y – 2xy2|(1,3) = –9,and L = fy|(1,3)= x3 – 2x2y|(1,3) = –5.Hence dz = M*dx + L* dy = –9(0.03) – 5(–0.06) = 0.03,
65. 65. Total DifferentialsExample B. Approximate the value of x3y – x2y2at (x, y) = (1.03, 2.94) using the differentials.Let’s set z = f(x,y) = x3y – x2y2. Select P = (1, 3)which is close to (1.03, 2.94) and easy to evaluate.At P = (1, 3), z = 3 – 9 = –6. To find an approximatevalue of f(1.03, 2.94), we calculate the z–value of(1.03, 2.94) on the tangent plane that touches (1, 3, 6).From (1, 3) to (1.03, 2.94), dx = 0.03 and dy = –0.06.The partials at (1, 3), M = fx|(1,3)= 3x2y – 2xy2|(1,3) = –9,and L = fy|(1,3)= x3 – 2x2y|(1,3) = –5.Hence dz = M*dx + L* dy = –9(0.03) – 5(–0.06) = 0.03,so f(1.03, 2.94) ≈ f(1, 3) + dz = –6 + 0.03 = –5.970.
66. 66. Total DifferentialsExample B. Approximate the value of x3y – x2y2at (x, y) = (1.03, 2.94) using the differentials.Let’s set z = f(x,y) = x3y – x2y2. Select P = (1, 3)which is close to (1.03, 2.94) and easy to evaluate.At P = (1, 3), z = 3 – 9 = –6. To find an approximatevalue of f(1.03, 2.94), we calculate the z–value of(1.03, 2.94) on the tangent plane that touches (1, 3, 6).From (1, 3) to (1.03, 2.94), dx = 0.03 and dy = –0.06.The partials at (1, 3), M = fx|(1,3)= 3x2y – 2xy2|(1,3) = –9,and L = fy|(1,3)= x3 – 2x2y|(1,3) = –5.Hence dz = M*dx + L* dy = –9(0.03) – 5(–0.06) = 0.03,so f(1.03, 2.94) ≈ f(1, 3) + dz = –6 + 0.03 = –5.970.The calculator answer of f(1.03, 2.94) ≈ –5.957.