**Problem 4:**Let $ T(r, s) = F(x(r, s), y(r, s)) $ be a differentiable function, where:- $ x = r + s $- $ y = \frac{r - s}{rs} $Find $ T_r(1, 1) $ and $ T_s(1, 1) $ given the following conditions:- $ F(2, 0) = 0 $- $ F(1, 1) = -3 $- $ F_x(2, 0) = -3 $- $ F_x(1, 1) = 5 $- $ F_y(2, 0) = -5 $- $ F_y(1, 1) = 11 $

Tr,s=Fxr,s,yr,s x=r+sy=r-srs∴∂x∂r=1∂x∂s=1∂y∂r=1sr-r+sr2 =1r2∂y∂s=1r-s-r-ss2…

Answered: T-S 4. Let T(r, s) = F(x(r, s), y(r,…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

**Problem 4:**

Let $ T(r, s) = F(x(r, s), y(r, s)) $ be a differentiable function, where:

- $ x = r + s $
- $ y = \frac{r - s}{rs} $

Find $ T_r(1, 1) $ and $ T_s(1, 1) $ given the following conditions:

- $ F(2, 0) = 0 $
- $ F(1, 1) = -3 $
- $ F_x(2, 0) = -3 $
- $ F_x(1, 1) = 5 $
- $ F_y(2, 0) = -5 $
- $ F_y(1, 1) = 11 $

Expert Solution

Step 1: Solution

$T (r, s) = F (x (r, s), y (r, s))$

$x = r + s y = \frac{r - s}{r s} ∴ \frac{\partial x}{\partial r} = 1 \frac{\partial x}{\partial s} = 1 \frac{\partial y}{\partial r} = \frac{1}{s} (\frac{r - r + s}{r^{2}}) = \frac{1}{r^{2}} \frac{\partial y}{\partial s} = \frac{1}{r} (\frac{- s - (r - s)}{s^{2}}) = - \frac{1}{s^{2}} T_{r} (r, s) = \frac{\partial}{\partial r} (F (x (r, s), y (r, s))) = F_{x} (x (r, s), y (r, s)) \frac{\partial x}{\partial r} + F_{y} (x (r, s), y (r, s)) \frac{\partial y}{\partial r} T_{r} (1, 1) = F_{x} (x (1, 1), y (1, 1)) {[\frac{\partial x}{\partial r}]}_{r = 1, s = 1} + F_{y} (x (1, 1), y (1, 1)) {[\frac{\partial y}{\partial r}]}_{r = 1, s = 1} = F_{x} (1 + 1, \frac{1 - 1}{1 \cdot 1}) {[\frac{\partial x}{\partial r}]}_{r = 1, s = 1} + F_{y} (1 + 1, \frac{1 - 1}{1 \cdot 1}) {[\frac{\partial y}{\partial r}]}_{r = 1, s = 1} = F_{x} (2, 0) \cdot 1 + F_{y} (2, 0) \cdot 1 = - 3 - 5 [∵ F_{x} (2, 0) = - 3, F_{y} (2, 0) = - 5 (g i v e n)] = - 8 T_{s} (1, 1) = F_{x} (x (1, 1), y (1, 1)) {[\frac{\partial x}{\partial s}]}_{r = 1, s = 1} + F_{y} (x (1, 1), y (1, 1)) {[\frac{\partial y}{\partial s}]}_{r = 1, s = 1} = F_{x} (1 + 1, \frac{1 - 1}{1 \cdot 1}) {[\frac{\partial x}{\partial s}]}_{r = 1, s = 1} + F_{y} (1 + 1, \frac{1 - 1}{1 \cdot 1}) {[\frac{\partial y}{\partial s}]}_{r = 1, s = 1} = F_{x} (2, 0) \cdot {[1]}_{r = 1 . s = 1} + F_{y} (2, 0) \cdot {[- \frac{1}{s^{2}}]}_{r = 1, s = 1} = - 3 \cdot 1 - 5 \cdot (- 1) = 2$