Q20P If u+v=x3-y3+4,u2-v2=x2y2+1, fin... [FREE SOLUTION]

Chapter 4: Q20P (page 210)

If $u + v = x^{3} - y^{3} + 4, u^{2} - v^{2} = x^{2} y^{2} + 1$ , find ${(\frac{鈭� u}{鈭� x})}_{y}, {(\frac{鈭� u}{鈭� x})}_{v}, {(\frac{鈭� x}{鈭� u})}_{y}, {(\frac{鈭� x}{鈭� u})}_{v}$ at $(x, y, u, v) = (2, - 1, 3, 2)$ .

Short Answer

Expert verified

The value of ${(\frac{鈭� u}{鈭� x})}_{y}$ is $\frac{4}{3}$ .

The value of ${(\frac{鈭� u}{鈭� x})}_{v}$ is $\frac{14}{5}$ .

The value of ${(\frac{鈭� x}{鈭� u})}_{y}$ is $\frac{3}{4}$ .

The value of ${(\frac{鈭� x}{鈭� u})}_{v}$ is $\frac{5}{14}$ .

Step by step solution

Given information.

Given $u^{2} + v^{2} = x^{3} - y^{3} + 4$ , $u^{2} - v^{2} = x^{2} y^{2} + 1$

Definition of partial differentiation.

Partial differentiation is defined as the process, in which find the partial derivative of a function.

In Partial differentiation, the function has more than one variable and find the partial derivative of a function with respect to one variable and keeping the other variable constant.

Find (∂u∂x)y.

Find the partial differentiation of function, $u^{2} + v^{2} = x^{3} - y^{3} + 4$ with respect to x.

$2 u \frac{鈭� u}{鈭� x} + 2 v \frac{鈭� v}{鈭� x} = 3 x^{2} - 3 y^{2} \frac{鈭� y}{鈭� x} . . . . . . . . . . . (1)$

Now find the partial differentiation of function, $u^{2} - v^{2} = x^{2} y^{2} + 1$ with respect to x.

$2 u \frac{鈭� u}{鈭� x} - 2 v \frac{鈭� v}{鈭� x} = 2 x y^{2} + 2 x^{2} y \frac{鈭� y}{鈭� x} . . . . . . . . (2)$

Add both equations and simplify it.

$4 u \frac{鈭� u}{鈭� x} = 3 x^{2} + 2 x y^{2} - 3 y^{2} \frac{鈭� y}{鈭� x} + 2 x^{2} y \frac{鈭� y}{d x} \frac{鈭� u}{鈭� x} = \frac{1}{u} [3 x^{2} + 2 x y^{2} - 3 y^{2} \frac{鈭� v}{鈭� x} + 2 x^{2} y \frac{鈭� y}{d x}]$

Now find ${(\frac{鈭� u}{鈭� x})}_{y}$ by keeping is constant.

${(\frac{鈭� u}{鈭� x})}_{y} = \frac{1}{4.3} (3 . 2^{2} + 2.2 . {(- 1)}^{2}) = \frac{1}{12} (12 + 4) = \frac{16}{12} = \frac{4}{3}$

Substitute the values of ${(\frac{鈭� u}{鈭� x})}_{y}$ is $\frac{4}{3}$ .

Find (∂u∂x)v.

From equation (1), find ${(\frac{鈭� u}{鈭� u})}_{v}$ by keeping v is a constant.

$2 u {(\frac{鈭� u}{鈭� x})}_{v} + 2 v . 0 = 3 x^{2} - 3 y^{2} \frac{鈭� y}{鈭� x} 2 u {(\frac{鈭� u}{鈭� x})}_{v} = 3 x^{2} - 3 y^{2} \frac{鈭� y}{鈭� x} . . . . . . . . . . . (3)$

Similarly from equation (2), find ${(\frac{鈭� u}{鈭� x})}_{v}$ by keeping v is a constant.

$2 u {(\frac{鈭� y}{鈭� x})}_{v} - 2 y . 0 = 2 x y^{2} + 2 x^{2} y \frac{鈭� y}{鈭� x} 2 u {(\frac{鈭� y}{鈭� x})}_{v} = 2 x y^{2} + 2 x^{2} y 2 u \frac{鈭� y}{鈭� x} . . . . . . . . . . . (4)$

From equation (3) and (4),

$3 x^{2} - 3 y^{2} \frac{鈭� y}{鈭� x} = 2 x y^{2} + 2 x^{2} y \frac{鈭� y}{鈭� x} - \frac{鈭� y}{鈭� x} (3 y^{2} + 2 x^{2} y) = 2 x y^{2} - 3 x^{2} \frac{鈭� y}{鈭� x} = \frac{3 x^{2} - 2 x y^{2}}{3 y^{2} 2 x^{2} y} . . . . . . . . . . . (5)$

Now substitute the value of $\frac{鈭� y}{鈭� x}$ from equation (5) into equation (3) and find ${(\frac{鈭� u}{鈭� x})}_{v}$ .

$2 u {(\frac{鈭� u}{鈭� x})}_{v} = 3 x^{2} - 3 y^{2} (\frac{3 x^{2} - 2 x y^{2}}{3 y^{2} + 2 x^{2} y}) 2 u {(\frac{鈭� u}{鈭� x})}_{v} = \frac{9 x^{2} y^{2} + 6 x^{2} y - 9 x^{2} y^{2} + 6 x y^{4}}{3 y^{2} + 2 x^{2} y} {(\frac{鈭� u}{鈭� x})}_{v} = \frac{1}{2 u} (\frac{9 x^{2} y^{2} + 6 x^{2} y - 9 x^{2} y^{2} + 6 x y^{4}}{3 y^{2} + 2 x^{2} y})$

Substitute the values $(x, y, u, v) = (2, - 1, 3, 2)$ .

${(\frac{鈭� u}{鈭� x})}_{v} = \frac{1}{2 u} (\frac{9 x^{2} y^{2} + 6 x^{2} y - 9 x^{2} y^{2} + 6 x y^{2}}{3 y^{2} + 2 x^{2} y}) = \frac{1}{6} (\frac{9 脳 4 脳 1 - 6 脳 16 脳 1 - 9 脳 4 脳 1 + 6 脳 2 脳 1}{3 脳 1 - 2 脳 4 脳 1}) = \frac{1}{6} (\frac{36 - 96 - 36 + 12}{3 - 8}) = \frac{1}{6} (\frac{- 84}{- 5}) = \frac{14}{5} Hence the value of {(\frac{鈭� u}{鈭� x})}_{v} is \frac{14}{5} .$

Find (∂x∂u)y.

Find the partial differentiation of function, $u^{2} + v^{2} = x^{3} - y^{3} + 4$ with respect to u.

$2 u + 2 v \frac{鈭� v}{鈭� u} = 3 x^{2} \frac{鈭� x}{鈭� u} - 3 y^{2} \frac{鈭� y}{鈭� u} . . . . . . . . . . (6)$

Now find the partial differentiation of function, $u^{2} - v^{2} = x^{2} y^{2} + 1$ with respect to u.

$2 u - 2 v \frac{鈭� v}{鈭� u} = 2 x y^{2} \frac{鈭� x}{鈭� u} + 2 x^{2} y \frac{鈭� y}{鈭� u} . . . . . . . . (7)$

Now find ${(\frac{鈭� x}{鈭� u})}_{y}$ from equation (6) by keeping y is constant.

$2 u + 2 v \frac{鈭� v}{鈭� u} = 3 x^{2} {(\frac{鈭� v}{鈭� u})}_{y} - 3 y^{2} . 0 {(\frac{鈭� x}{鈭� u})}_{y} = \frac{1}{3 x^{2}} (2 u + 2 v \frac{鈭� v}{鈭� u}) . . . . . . . . . (8)$

Now find ${(\frac{鈭� x}{鈭� u})}_{y}$ from equation (7) by keepingy is constant.

$2 u - 2 v \frac{鈭� v}{鈭� u} = 2 x y^{2} {(\frac{鈭� v}{鈭� u})}_{y} + 2 x y^{2} . 0 2 u - 2 v \frac{鈭� v}{鈭� u} = 2 x y^{2} {(\frac{鈭� v}{鈭� u})}_{y} {(\frac{鈭� x}{鈭� u})}_{y} = \frac{1}{2 x y^{2}} (2 u - 2 v \frac{鈭� v}{鈭� u}) . . . . . . . . . (9)$

From equation (8) and (9), find the value of $\frac{鈭� y}{鈭� u}$ .

$\frac{1}{3 x^{2}} (2 u + 2 v \frac{鈭� v}{鈭� u}) = \frac{1}{2 x y^{2}} (2 u - 2 v \frac{鈭� v}{鈭� u}) (2 u + 2 v \frac{鈭� v}{鈭� u}) = \frac{3 x^{2}}{2 x y^{2}} (2 u - 2 v \frac{鈭� v}{鈭� u}) 2 v \frac{鈭� v}{鈭� u} (1 + \frac{3 x}{2 y^{2}}) = 2 u (\frac{3 x}{2 y^{2}} - 1) \frac{鈭� v}{鈭� u} = (\frac{u (\frac{3 x}{2 y^{2}} - 1)}{v (1 + \frac{3 x}{2 y^{2}})})$

Now substitute the value of $\frac{鈭� v}{鈭� u}$ in to equation (8).

${(\frac{鈭� x}{鈭� u})}_{y} = \frac{1}{3 x^{2}} (2 u + 2 v . \frac{u (\frac{3 x}{2 y^{2}} - 1)}{v (1 + \frac{3 x}{2 y^{2}})}) = \frac{1}{3 x^{2}} (2 u + 2 v . \frac{(\frac{3 x}{2 y^{2}} - 1)}{(1 + \frac{3 x}{2 y^{2}})}) = \frac{2 u}{3 x^{2}} (1 + \frac{(\frac{3 x}{2 y^{2}} - 1)}{(1 + \frac{3 x}{2 y^{2}})})$

Substitute the values $(x, y, u, v) = (2, - 1, 3, 2)$ .

${(\frac{鈭� x}{鈭� v})}_{y} = \frac{2 u}{3 x^{2}} (1 + \frac{(\frac{3 x}{2 y^{2}} - 1)}{(1 + \frac{3 x}{2 y^{2}})}) = \frac{6}{12} (1 + \frac{(\frac{6}{2} - 1)}{(1 + \frac{6}{2})}) = \frac{1}{2} (1 + \frac{2}{4}) = \frac{1}{2} 脳 \frac{3}{2} = \frac{3}{2}$

Hence the value of ${(\frac{鈭� x}{鈭� v})}_{y}$ is $\frac{3}{4}$ .

Find (∂x∂u)v .

The partial differentiation of function, $u^{2} + v^{2} = x^{3} - y^{3} + 4$ with respect tou is as follows:

$2 u + 2 v \frac{鈭� v}{鈭� u} = 3 x^{2} \frac{鈭� x}{鈭� u} - 3 y^{2} \frac{鈭� y}{鈭� u} . . . . . . . . . . (10)$

The partial differentiation of function, $u^{2} - v^{2} = x^{2} y^{2} + 1$ with respect to u. Is as follows:

$2 u - 2 v \frac{鈭� v}{鈭� u} = 2 x y^{2} \frac{鈭� x}{鈭� u} + 2 x^{2} y \frac{鈭� y}{鈭� u} . . . . . . . . (11)$

Now find ${(\frac{鈭� x}{鈭� u})}_{v}$ from equation (10) by keepingv is constant.

$2 u + 2 v . 0 = 3 x^{2} {(\frac{鈭� x}{鈭� u})}_{v} - 3 y^{2} \frac{鈭� y}{鈭� u} 3 x^{2} {(\frac{鈭� x}{鈭� u})}_{v} = 3 y^{2} \frac{鈭� y}{鈭� u} + 2 u {(\frac{鈭� x}{鈭� u})}_{v} = \frac{1}{3 x^{2}} (3 x^{2} \frac{鈭� y}{鈭� u} + 2 u) . . . . . . . . . . . . (12)$

Now find ${(\frac{鈭� x}{鈭� u})}_{v}$ from equation (11) by keeping v is constant.

$2 u + 2 v . 0 = 2 x y^{2} {(\frac{鈭� x}{鈭� u})}_{v} + 2 x^{2} y \frac{鈭� y}{鈭� u} 2 x y^{2} {(\frac{鈭� x}{鈭� u})}_{v} = 2 u - 2 x^{2} y \frac{鈭� y}{鈭� u} {(\frac{鈭� x}{鈭� u})}_{v} = \frac{1}{2 x y^{2}} [2 u - 2 x^{2} y \frac{鈭� y}{鈭� u}] . . . . . . . . . . . . (13)$

From equation (12) and (13), find the value of $\frac{鈭� y}{鈭� u}$ .

$\frac{1}{3 x^{2}} (3 y^{2} \frac{鈭� y}{鈭� u} + 2 u) = \frac{1}{2 x y^{2}} [2 u - 2 x^{2} y \frac{鈭� v}{鈭� u}] 3 y^{2} \frac{鈭� y}{鈭� u} + 2 u = \frac{3 x^{2}}{2 x y^{2}} [2 u - 2 x^{2} y \frac{鈭� y}{鈭� u}] \frac{鈭� y}{鈭� u} (3 y^{2} + \frac{6 x^{2} y}{2 x y^{2}}) = \frac{6 x^{2} y}{2 x y^{2}} - 2 u \frac{鈭� y}{鈭� u} (3 y^{2} + \frac{3 x^{2}}{y}) = u (\frac{3 x}{y^{2}} - 2) \frac{鈭� y}{鈭� u} \frac{u (\frac{3 x}{y^{2}} - 2)}{(3 y^{2} \frac{3 x^{3}}{y})}$

Now substitute the value of $\frac{鈭� y}{鈭� u}$ in to equation (12).

${(\frac{鈭� x}{鈭� u})}_{v} = \frac{1}{3 x^{2}} (3 y^{2} \frac{u (\frac{3 x}{y^{2}} - 2)}{(3 y^{2} + \frac{3 x^{3}}{y})} + 2 u) = \frac{u y^{2} (\frac{3 x}{y^{2}} - 2)}{x^{2} (3 y^{2} \frac{3 x^{3}}{y})} + \frac{2 u}{3 x^{2}}$

Substitute the values $(x, y, u, v) = (2, - 1, 3, 2)$ .

${(\frac{鈭� x}{鈭� u})}_{v} = \frac{u y^{2} (\frac{3 x^{2}}{y^{2}} - 2)}{x^{2} (3 y^{2} + \frac{3 x^{3}}{y})} + \frac{2 u}{3 x^{2}} = \frac{3 (\frac{6}{1} - 2)}{4 (3 + \frac{24}{- 1})} + \frac{6}{12}$

Now simplify the expression.

${(\frac{鈭� x}{鈭� u})}_{v} = \frac{12}{- 84} + \frac{1}{2} = - \frac{1}{7} + \frac{1}{2} = \frac{- 2 + 7}{14} = \frac{5}{14}$

Hence the value of ${(\frac{鈭� x}{鈭� u})}_{v}$ is $\frac{5}{14}$ .

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If $u + v = x^{3} - y^{3} + 4, u^{2} - v^{2} = x^{2} y^{2} + 1$ , find ${(\frac{鈭� u}{鈭� x})}_{y}, {(\frac{鈭� u}{鈭� x})}_{v}, {(\frac{鈭� x}{鈭� u})}_{y}, {(\frac{鈭� x}{鈭� u})}_{v}$ at $(x, y, u, v) = (2, - 1, 3, 2)$ .

Short Answer

Step by step solution

Given information.

Definition of partial differentiation.

Find (∂u∂x)y.

Find (∂u∂x)v.

Find (∂x∂u)y.

Find (∂x∂u)v .

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If u+v=x3-y3+4,u2-v2=x2y2+1, find (鈭�u鈭�x)y,(鈭�u鈭�x)v,(鈭�x鈭�u)y,(鈭�x鈭�u)vat(x,y,u,v)=(2,-1,3,2).

Short Answer

Step by step solution

Given information.

Definition of partial differentiation.

Find (∂u∂x)y.

Find (∂u∂x)v.

Find (∂x∂u)y.

Find (∂x∂u)v .

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Particle Model of Matter

Waves Physics

Circular Motion and Gravitation

Fluids

Rotational Dynamics

Classical Mechanics

Study anywhere. Anytime. Across all devices.

If $u + v = x^{3} - y^{3} + 4, u^{2} - v^{2} = x^{2} y^{2} + 1$ , find ${(\frac{鈭� u}{鈭� x})}_{y}, {(\frac{鈭� u}{鈭� x})}_{v}, {(\frac{鈭� x}{鈭� u})}_{y}, {(\frac{鈭� x}{鈭� u})}_{v}$ at $(x, y, u, v) = (2, - 1, 3, 2)$ .