d) f(r,y) = r= 0, y = 1, dr = 0.01 and dy = 0.03 %3D %3D for f.(r, y), let u(r, y) = 4x-y and v(x, y) = 5x + 4y. then, u (r, y) = (i) 4x? (ii) 4 (iii) –1 and v(r, y) = (i) 5x? (ii) 5 (iii) 3+ 4x and so v(z, y) - u,(r. y) – u(x, y) v,(r, y)_ (5x+ 4y)(4) – (4x – y)(5) (5x + 4y)? fr(x, y) = [v(x, y) 21 21y (5z+4y) 20 (i) (5z+4y) (ii) (iii) (5z+4y) for fy(x, y), let u(r, y) = 4.r - y and v(r, y) = 5x + 4y. then, u,(r, y) = (i) 4x? (ii) 4 (iii) –1 and vy(r, y) = (i) 5x? (ii) 5 (iii) 4 and so %3D v(x, y) - u,(x, y) – u(x,y) - v,(r. y) _ (5x +4y)(-1) – (4r – y)(4) (5r+4y)2 fy(r, y) = [v(x, y)* (i) 21 (5x+4y) 21y (ii) (5z+4y) (iii) -21z (5z+4y) dz = f.(x,y) dr + f,(x, y) dy 21y -21r (5x + 4y)?) dr + (5.r + 4y)2 dy 21(1) (5(0) + 4(1))² (i) 0.011 (ii) 0.012 (iii) 0.013 -21(0) (0.01) + (0.03) = (5(0) + 4(1))? for f.(r, y), let f[9(x)] = (x² + y²) 1 (2x) = 2 + y? (1) T (ii) V+y? (iii) 3V3z+y for f,(r, y), 1 (2y) 3D 2/3r + y* fy (1) (i) 3z (iii) 3/3z+y?
d) f(r,y) = r= 0, y = 1, dr = 0.01 and dy = 0.03 %3D %3D for f.(r, y), let u(r, y) = 4x-y and v(x, y) = 5x + 4y. then, u (r, y) = (i) 4x? (ii) 4 (iii) –1 and v(r, y) = (i) 5x? (ii) 5 (iii) 3+ 4x and so v(z, y) - u,(r. y) – u(x, y) v,(r, y)_ (5x+ 4y)(4) – (4x – y)(5) (5x + 4y)? fr(x, y) = [v(x, y) 21 21y (5z+4y) 20 (i) (5z+4y) (ii) (iii) (5z+4y) for fy(x, y), let u(r, y) = 4.r - y and v(r, y) = 5x + 4y. then, u,(r, y) = (i) 4x? (ii) 4 (iii) –1 and vy(r, y) = (i) 5x? (ii) 5 (iii) 4 and so %3D v(x, y) - u,(x, y) – u(x,y) - v,(r. y) _ (5x +4y)(-1) – (4r – y)(4) (5r+4y)2 fy(r, y) = [v(x, y)* (i) 21 (5x+4y) 21y (ii) (5z+4y) (iii) -21z (5z+4y) dz = f.(x,y) dr + f,(x, y) dy 21y -21r (5x + 4y)?) dr + (5.r + 4y)2 dy 21(1) (5(0) + 4(1))² (i) 0.011 (ii) 0.012 (iii) 0.013 -21(0) (0.01) + (0.03) = (5(0) + 4(1))? for f.(r, y), let f[9(x)] = (x² + y²) 1 (2x) = 2 + y? (1) T (ii) V+y? (iii) 3V3z+y for f,(r, y), 1 (2y) 3D 2/3r + y* fy (1) (i) 3z (iii) 3/3z+y?
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter2: Functions And Graphs
Section2.6: Proportion And Variation
Problem 22E: Find the constant of proportionality. z is directly proportional to the sum of x and y. If x=2 and...
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![L, r = 0, y = 1, dr = 0.01 and dy = 0.03
Sz+4y
4r-y
(d) f(x, y) =
%3D
%3D
for f.(r, y), let u(x, y) = 4x- y and v(x, y) = 5x + 4y.
then, u (r,y) = (i) 4x? (ii) 4 (iii) –1
and v(r, y) = (i) 5æ? (ii) 5 (iii) 3+ 4a
and so
%3D
v(x, y) - u,(x, y) – u(r, y) v,(r, y)_ (5x + 4y)(4) – (4x – y)(5)
(5x + 4y)2
fa(r, y) =
[v(x, y)]*
21
21y
(5z+4y)
20
(i)
(5a+4y)2
(ii)
(iii) T6z+4y)²
for fy(x, y), let u(r, y) = 4x - y and v(r, y) = 5x + 4y.
then, u,(r, y) = (i) 4x? (ii) 4 (iii) –1
and v, (x, y) = (i) 5æ? (ii) 5 (iii) 4
and so
%3D
%3D
v(a, y) - u,(x, y) – u(r, y)- v,(r, y)
[v(x, y)]*
(5.r + 4y)(-1) – (4.r – y)(4)
(5x + 4y)2
fy(x, y) =
(i)
21
(5æ+4y)"
(ii)
21y
(5a+4y)
21a
(iii) 6z+4y)
dz = fa(x, y) dr + f,(x, y) dy
21y
-21.r
dr +
(5x + 4y)2,
dy
(5x + 4y)2
-21(0)
21(1)
(5(0) + 4(1))2
(0.01) +
(0.03) =
(5(0) + 4(1))?
(i) 0.011 (ii) 0.012 (iii) 0.013
for f.(r, y), let f [g(x)] = (x² + y?).
1
f.
(2r) =D
2VT + y
(1) (i)
(ii)
3V3a+y?
a²+y?
for f,(r, y),
1
(2y) =
2/3r + y?
fy
(1) (i) (i)
3z
3/3r+y?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3d040c1d-23ed-4c39-83a8-90840b9e659f%2F340efcc7-5c44-43ff-8c41-c4067b01d884%2Fvb7ggz_processed.jpeg&w=3840&q=75)
Transcribed Image Text:L, r = 0, y = 1, dr = 0.01 and dy = 0.03
Sz+4y
4r-y
(d) f(x, y) =
%3D
%3D
for f.(r, y), let u(x, y) = 4x- y and v(x, y) = 5x + 4y.
then, u (r,y) = (i) 4x? (ii) 4 (iii) –1
and v(r, y) = (i) 5æ? (ii) 5 (iii) 3+ 4a
and so
%3D
v(x, y) - u,(x, y) – u(r, y) v,(r, y)_ (5x + 4y)(4) – (4x – y)(5)
(5x + 4y)2
fa(r, y) =
[v(x, y)]*
21
21y
(5z+4y)
20
(i)
(5a+4y)2
(ii)
(iii) T6z+4y)²
for fy(x, y), let u(r, y) = 4x - y and v(r, y) = 5x + 4y.
then, u,(r, y) = (i) 4x? (ii) 4 (iii) –1
and v, (x, y) = (i) 5æ? (ii) 5 (iii) 4
and so
%3D
%3D
v(a, y) - u,(x, y) – u(r, y)- v,(r, y)
[v(x, y)]*
(5.r + 4y)(-1) – (4.r – y)(4)
(5x + 4y)2
fy(x, y) =
(i)
21
(5æ+4y)"
(ii)
21y
(5a+4y)
21a
(iii) 6z+4y)
dz = fa(x, y) dr + f,(x, y) dy
21y
-21.r
dr +
(5x + 4y)2,
dy
(5x + 4y)2
-21(0)
21(1)
(5(0) + 4(1))2
(0.01) +
(0.03) =
(5(0) + 4(1))?
(i) 0.011 (ii) 0.012 (iii) 0.013
for f.(r, y), let f [g(x)] = (x² + y?).
1
f.
(2r) =D
2VT + y
(1) (i)
(ii)
3V3a+y?
a²+y?
for f,(r, y),
1
(2y) =
2/3r + y?
fy
(1) (i) (i)
3z
3/3r+y?
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