Question

1. If u = f(x,y) be a function of two independent variables x and y, then ∂u/∂y is equal to
a) lim△x→0f(x+△x,y)−f(x,y)△x
b) lim△y→0f(x,y+△y)−f(x,y)△y
c) limy→0f(x,y+△y)−f(x,y)△y
d) lim△y→0f(x+△x,y)−f(x,y)△x
2. Find the total differential of the function
u=x2y−3y

a)2x+(x2−3)dy
b) 2xydx+x2dy
c) 2xydx+(x2−3)dy
d) 2xydx+(x3−2)dy
3. The total differential du of a function u (x, y) = 0 is defined as

a) ∂u∂xdx+∂u∂xdy=0
b) ∂u∂xdy+∂u∂xdy=0
c) ∂u∂xdx+∂u∂ydx=0
d)∂u∂xdx+∂u∂ydy=0

Answer 1

Answer on Question #61735 – Math – Differential Equations

Question #1

If $u = f(x,y)$ be a function of two independent variables $x$ and $y$, then $\partial u / \partial y$ is equal to

a) $\lim \Delta x \to 0 \pmod{x + \Delta x, y} - f(x, y) / \Delta x$

b) $\lim \Delta y \to 0 \pmod{x, y + \Delta y} - f(x, y) / \Delta y$

c) $\lim y \to 0 \pmod{x, y + \Delta y} - f(x, y) / \Delta y$

d) $\lim \Delta y \to 0 \pmod{x + \Delta x, y} - f(x, y) / \Delta x$

Solution

$\frac{\partial u}{\partial y} = \lim_{\Delta y \to 0} \frac{f(x, y + \Delta y) - f(x, y)}{\Delta y}$ according to the definition of the partial derivative.

**Answer:** b) $\lim_{\Delta y \to 0} \frac{f(x,y + \Delta y) - f(x,y)}{\Delta y}$.

Question #2

Find the total differential of the function:

$$u = x^2 y - 3 y.$$

a) $2x + (x^2 - 3)dy$

b) $2xydx + x^2 dy$

c) $2xydx + (x^2 - 3)dy$

d) $2xydx + (x^3 - 2)dy$

Solution

Let $x$ and $y$ be independent variables. Then the total differential of the function $u = f(x, y)$ is calculated by the following formula:

$$du = df(x, y) = \frac{\partial f(x, y)}{\partial x} dx + \frac{\partial f(x, y)}{\partial y} dy,$$

where $\frac{\partial f(x,y)}{\partial x}$ and $\frac{\partial f(x,y)}{\partial y}$ are partial derivatives of the function $f(x,y)$ with respect to $x$ and $y$ respectively.

We hold $y$ fixed and allow $x$ to vary, then

$$\frac{\partial f(x, y)}{\partial x} = \frac{\partial}{\partial x} (x^2 y - 3 y) = \frac{\partial}{\partial x} (x^2 y) - \frac{\partial}{\partial x} (3 y) = y \frac{\partial}{\partial x} (x^2) - (3 y) \frac{\partial}{\partial x} (1) = 2 x y - 0 = 2 x y.$$

We hold $x$ fixed and allow $y$ to vary, then

$$\frac {\partial f (x , y)}{\partial y} = \frac {\partial}{\partial y} (x ^ {2} y - 3 y) = \frac {\partial}{\partial y} (x ^ {2} y) - \frac {\partial}{\partial y} (3 y) = x ^ {2} \frac {\partial}{\partial y} (y) - 3 \frac {\partial}{\partial y} (y) = x ^ {2} - 3.$$

Thus,

$$\mathrm {d u} = \mathrm {d f} (x, y) = \frac {\partial f (x , y)}{\partial x} d x + \frac {\partial f (x , y)}{\partial y} d y = 2 x y \mathrm {d x} + (x ^ {2} - 3) \mathrm {d y}.$$

Answer: c) $2xydx + (x^{2} - 3)dy$.

Question #3

The total differential du of a function $\mathbf{u}(\mathbf{x},\mathbf{y}) = 0$ is defined as ...

a) $\partial \mathbf{u} / \partial \mathbf{x}\mathrm{d}\mathbf{x} + \partial \mathbf{u} / \partial \mathbf{x}\mathrm{d}\mathbf{y} = 0$

b) $\partial \mathbf{u} / \partial \mathbf{x}\mathrm{d}\mathbf{y} + \partial \mathbf{u} / \partial \mathbf{x}\mathrm{d}\mathbf{y} = 0$

c) $\partial \mathbf{u} / \partial \mathbf{x}\mathrm{d}\mathbf{x} + \partial \mathbf{u} / \partial \mathbf{y}\mathrm{d}\mathbf{x} = 0$

d) $\partial \mathbf{u} / \partial \mathbf{x}\mathrm{d}\mathbf{x} + \partial \mathbf{u} / \partial \mathbf{y}\mathrm{d}\mathbf{y} = 0$

Solution

Let $x$ and $y$ be independent variables. Then the total differential of the function $\mathbf{u} = \mathbf{f}(x, y)$ is calculated by the following formula:

$$\mathrm {d u} = \mathrm {d f} (x, y) = \frac {\partial f (x , y)}{\partial x} d x + \frac {\partial f (x , y)}{\partial y} d y,$$

where $\frac{\partial f(x,y)}{\partial x}$ and $\frac{\partial f(x,y)}{\partial y}$ are partial derivatives of the function $f(x,y)$ with respect to $x$ and $y$ respectively.

Thus, the correct answer is

$$\frac {\partial f (x , y)}{\partial x} d x + \frac {\partial f (x , y)}{\partial y} d y = 0.$$

Answer: d) $\frac{\partial f(x,y)}{\partial x} dx + \frac{\partial f(x,y)}{\partial y} dy = 0.$

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