Let γ be a real constant with y² 1 for the parametric functional S[x, y] = fjª dt [√ಠ+ 2y àÿ + ÿj² − X(xÿ − ±y)], \>0, 0 - - with the boundary conditions x(0) = y(0) = 0, x(1) = R > 0 and y(1) = 0. (1 - y²)x' = D-X (1-2) y' = C + Y where D = 2(c-yd), X = 2λ(yx + y), where C = 2(d- yc), Y = 2λ(x + yy), where dx dy x' = y' y = ds' ds' x²+Y2-2yXY - 4c(1-2)x + 4d(1-2)Y +C² + D² + 2xCD = (1 − y²)². - C² + 2xCD + D² = (1 − y²)² -
Let γ be a real constant with y² 1 for the parametric functional S[x, y] = fjª dt [√ಠ+ 2y àÿ + ÿj² − X(xÿ − ±y)], \>0, 0 - - with the boundary conditions x(0) = y(0) = 0, x(1) = R > 0 and y(1) = 0. (1 - y²)x' = D-X (1-2) y' = C + Y where D = 2(c-yd), X = 2λ(yx + y), where C = 2(d- yc), Y = 2λ(x + yy), where dx dy x' = y' y = ds' ds' x²+Y2-2yXY - 4c(1-2)x + 4d(1-2)Y +C² + D² + 2xCD = (1 − y²)². - C² + 2xCD + D² = (1 − y²)² -
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter9: Multivariable Calculus
Section9.2: Partial Derivatives
Problem 25E
Related questions
Question
Show that C=-λR.
![Let γ be a real constant with y² 1 for the parametric functional
S[x, y] = fjª dt [√ಠ+ 2y àÿ + ÿj² − X(xÿ − ±y)], \>0,
0
-
-
with the boundary conditions x(0) = y(0) = 0, x(1) = R > 0 and y(1) = 0.
(1 - y²)x' = D-X
(1-2) y' = C + Y
where D = 2(c-yd), X = 2λ(yx + y),
where C = 2(d- yc), Y = 2λ(x + yy),
where
dx
dy
x'
=
y'
y =
ds'
ds'
x²+Y2-2yXY - 4c(1-2)x + 4d(1-2)Y
+C² + D² + 2xCD = (1 − y²)².
-
C² + 2xCD + D² = (1 − y²)²
-](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa6c8ed7d-75cc-4e27-869e-3ad6a1efc0b4%2Fb8e275ba-b18a-449b-96e3-1523c30baee3%2F4zf2mli_processed.png&w=3840&q=75)
Transcribed Image Text:Let γ be a real constant with y² 1 for the parametric functional
S[x, y] = fjª dt [√ಠ+ 2y àÿ + ÿj² − X(xÿ − ±y)], \>0,
0
-
-
with the boundary conditions x(0) = y(0) = 0, x(1) = R > 0 and y(1) = 0.
(1 - y²)x' = D-X
(1-2) y' = C + Y
where D = 2(c-yd), X = 2λ(yx + y),
where C = 2(d- yc), Y = 2λ(x + yy),
where
dx
dy
x'
=
y'
y =
ds'
ds'
x²+Y2-2yXY - 4c(1-2)x + 4d(1-2)Y
+C² + D² + 2xCD = (1 − y²)².
-
C² + 2xCD + D² = (1 − y²)²
-
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