Engineering Mathematics Test 3

Concept:

M dx + N dy = 0, will be exact differential equation if:

\(\frac{{\partial {\rm{M}}}}{{\partial {\rm{y}}}} = \frac{{\partial {\rm{N}}}}{{\partial {\rm{x}}}}\)

Where M and N are functions of x and y.

Given:

Equation (3xy2 + n2 x2y) dx + (nx3 + 3x2y) dy = 0, x ≠ 0

Here M = 3xy² + n² x²y

N = nx³ + 3x²y

Given equation is exact, thus:

6xy + n² x² = 3nx² + 6xy

n² x² = 3nx²

n = 3

Given:

u = log (x² + y²) where x + y + xy = 4

\(du = \frac{{\partial u}}{{\partial x}}dx + \frac{{\partial u}}{{\partial y}}dy\)

\(\therefore \frac{{du}}{{dx}} = \frac{{\partial u}}{{\partial x}}\frac{{dx}}{{dx}} + \frac{{\partial u}}{{\partial y}}\frac{{dy}}{{dx}}\)

Also,

\(\frac{{dy}}{{dx}} = \frac{{ - \partial f/\partial x}}{{\partial f/\partial y}}\)

Let,

f(x, y) = x + y + xy = 4

then

\(\frac{{\partial f}}{{\partial x}} = 1 + y\;and\frac{{\partial f}}{{\partial y}} = 1 + x\)

\(\therefore \frac{{du}}{{dx}} = \frac{{2x}}{{\left( {{x^2} + {y^2}} \right)}} + \frac{{2y}}{{\left( {{x^2} + {y^2}} \right)}}\left[ {\frac{{ - 1\left( {1 + y} \right)}}{{\left( {1 + x} \right)}}} \right]\)

\(\therefore {\left( {\frac{{du}}{{dx}}} \right)_{\left( {1,1} \right)}} = \frac{2}{{\left( {1 + 1} \right)}} + \frac{2}{{\left( {1 + 1} \right)}}\left\{ { - \frac{2}{2}} \right\} = 0\)

f(x) = sin x - cos x

f’(x) = cos x + sin x

f’(c) = 0 ⇒ cos c + sin c = 0

tan c = -1

\(C = - \frac{\pi }{4},\frac{{3\pi }}{4},\frac{{7\pi }}{4}\)

Since, C ϵ [0, π]

\(\therefore C = \frac{{3\pi }}{4}\)

Explanation:

Given:

f(x) = e^x, g(x) = e^-x

Now,

derivative of f(x) i.e. f^’(x) = e^x

derivative of g(x) i.e. g^’(x) = - e^x

range is given as [2, 3], that is, a = 2 and b = 3

here, f(x) and g(x) are differentiable and continuous and derivative of g(x) is not equal to 0.

They are satisfying the conditions of Cauchy’s mean value theorem.

Now,

Consider it is in interval of [a, b].

\(\frac{{f'\left( c \right)}}{{g'\left( c \right)}} = \frac{{f\left( b \right) - f\left( a \right)}}{{g\left( b \right) - g\left( a \right)}}\)

\(\frac{{{e^c}}}{{ - {e^{ - c}}}} = \frac{{{e^b} - {e^a}}}{{{e^{ - b}} - \;{e^{ - a}}}}\)

\(- {e^{2c}} = \frac{{{e^3} - {e^2}}}{{{e^{ - 3}} - \;{e^{ - 2}}}} - \; = \; - {e^{3 + 2}}\)

Explanation:

Given:

 \(u = x\log xy\)

\({x^3} + {y^3} + 3xy = 1\)

we know that

\(\frac{{\partial u}}{{\partial x}} = \frac{{\partial u}}{{\partial x}} + \frac{{\partial u}}{{\partial y}}\frac{{\partial y}}{{\partial x}}\)

\(\frac{{\partial u}}{{\partial x}} = x.\frac{1}{{xy}}.y + \log xy = 1 + \log xy\)

\(\frac{{\partial u}}{{\partial y}} = x.\frac{1}{{xy}}.x = \frac{x}{y}\)

\( \Rightarrow 3{x^2} + 3{y^2}\frac{{\partial y}}{{\partial x}} + 3\left( {x\frac{{\partial y}}{{\partial x}} + y.1} \right) = 0\)

\( \Rightarrow \frac{{\partial y}}{{\partial x}} = - \left( {\frac{{{x^2} + y}}{{{y^2} + x}}} \right)\)

\( \Rightarrow \frac{{\partial u}}{{\partial x}} = \left( {1 + \log xy} \right) + \frac{x}{y}\left\{ { - \left( {\frac{{{x^2} + y}}{{{y^2} + x}}} \right)} \right\}\)

\( \frac{{\partial u}}{{\partial x}} =\left( {1 + \log xy} \right) - \frac{x}{y}\left( {\frac{{{x^2} + y}}{{{y^2} + x}}} \right)\)

Explanation:

\(f\left( x \right) = \frac{1}{x},\;g\left( x \right) = \frac{1}{{{x^2}}}\)

f(x) and g(x) are continuous in [4, 6]

f(x) and g(x) are differentiable in [4, 6]

\(g'\left( x \right) = \frac{{ - 2}}{{{x^3}}}\)

g’(x) ≠ 0 in (4, 6)

Now, according to Cauchy’s mean value theorem

\(\frac{{f'\left( c \right)}}{{g'\left( c \right)}} = \frac{{f\left( b \right) - f\left( a \right)}}{{g\left( b \right) - g\left( a \right)}}\)

\(\begin{array}{*{20}{c}} {f'\left( x \right) = \frac{{ - 1}}{{{x^2}}}}&{f'\left( b \right) = \frac{{ - 2}}{{{x^3}}}} \end{array}\)

\(f\left( a \right) = f\left( 4 \right) = \frac{1}{4}\;\)

\(f\left( b \right) = f\left( 6 \right) = \frac{1}{6}\)

\(\begin{array}{l} g\left( a \right) = g\left( 4 \right) = \frac{1}{{16}}\\ g\left( b \right) = g\left( 6 \right) = \frac{1}{{36}} \end{array}\)

\(\begin{array}{l} \begin{array}{*{20}{c}} {f'\left( x \right) = \frac{{ - 1}}{{{c^2}}}}&{g'\left( c \right) = \frac{{ - 2}}{{{c^3}}}} \end{array}\\ \Rightarrow \frac{{\frac{{ - 1}}{{{c^2}}}}}{{\frac{{ - 2}}{{{c^3}}}}} = \frac{{\frac{1}{6} - \frac{1}{4}}}{{\frac{1}{{36}} - \frac{1}{{16}}}} = 2.4 \end{array}\)

∴ c = 4.8

Explanation:

Given:

Y = (A + Bx) e^3x is solution of

\(\frac{{{d^2}y}}{{d{x^2}}} = a\frac{{dy}}{{dx}} + by\)

Now,

Y = (A + Bx) e^3x

Ye^-3x = A + Bx

Differentiating,

Y(-3)e^-3x + e^-3x y’ = B

Differentiating again,

Y(9)e^-3x + e^-3x(-3x)y’ + e^-3xy’’ + y’(-3)e^-3x = 0

y’’ – 6y’ + 9y = 0

∴ y’’ = 6y’ – 9y

Now,

Comparing with given solution,

y’’ = ay’ + by

We get,

a = 6; b = -9

∴ a – b = 6 – (-9) = 15

Explanation:

f(x, y) = x³y² (1 - x - y)

\(\frac{{\partial f}}{{\partial x}} = 3{x^2}{y^2} - 4{x^3}{y^2} - 3{x^2}{y^3}\)

\(\frac{{\partial f}}{{\partial y}} = 2{x^3}y - 2{x^4}y - 3{x^3}{y^2}\)

\(\frac{{{\partial ^2}f}}{{\partial {x^2}}} = r = 6x{y^2} - 12{x^2}{y^2} - 6x{y^3}\)

\(\frac{{{\partial ^2}f}}{{\partial x \cdot \partial y}} = s = 6{x^2}y - 8{x^3}y - 9{x^2}{y^2}\)

\(\frac{{{\partial ^2}f}}{{\partial {y^2}}} = t = 2{x^3} - 2{x^4} - 6{x^3}y\)

Now,

Put \(\frac{{\partial f}}{{\partial x}} = 0,\) we get x²y² (3 - 4x - 3y) = 0

\(\frac{{\partial f}}{{\partial y}} = 0,\)

We get x³y (2 - 2x - 3y) = 0

Solving these two, we get stationary points \( \to \left( {\frac{1}{2},\frac{1}{3}} \right)\;\& \;\left( {0,\;0} \right)\) 

Now,

rt - s² = x⁴y² [12 (1 - 2x - y)(1 - x - 3y) - (6 - 8x - 9y)²]

Since x, y ϵ (0, ∞)

Thus, for \(\left( {\frac{1}{2},\frac{1}{3}} \right)\) 

\(rt - {s^2} = \frac{1}{{16}}\frac{1}{9}\left[ {12\left( { - \frac{1}{3}} \right)\left( { - \frac{1}{2}} \right) - {{\left( {6 - 4 - 3} \right)}^2}} \right]\) 

\(rt - {s^2} = \frac{1}{{14}} > 0\)

\(\& \;r = 6\left( {\frac{1}{2} \cdot \frac{1}{9} - \frac{2}{4} \cdot \frac{1}{9} - \frac{1}{2} \cdot \frac{1}{{27}}} \right) = - \frac{1}{9} < 0\)

Hence, f(x, y) has a maxima at \(\left( {\frac{1}{2},\frac{1}{3}} \right)\) 

\(\therefore {\rm{Maximum\;value}} = \frac{1}{8} \cdot \frac{1}{9}\left( {1 - \frac{1}{2} - \frac{1}{3}} \right) = \frac{1}{{432}}\)

Explanation:

f(x) = sin x + cos 2x in [0, 2π]

f’(x) = cos x – 2 sin 2x

Since sin2x = 2 sinx cosx, the above can be written as:

f'(x) = cos x [1 – 4 sinx]

Now,

Equating this to zero, we solve for x as:

f'(x) = cos x [1 – 4 sinx] = 0

⇒ cos x = 0; 1 – 4sinx = 0

For cosx = 0,

⇒ x = π/2, 3π/2 in [0, 2π]

Now,

Also, for sin x = 1/4, f'(x) = 0.

Also f’(x) exists for all x in [0, 2π]

Checking for all the points in the defined range, we get:

Now,

f(0) = 1; f(2π) = 1; f(π/2) = 0; f(3π/2) = -2

At sinx = 1/4 , f(x) = 1/2 + (1 – 2(1/4)²) = 9/8

∴ The maximum and minimum values of f(x) are 9/8 and -2 respectively, and they occur at:

 \(sinx = \frac{1}{4}\;and\;x = \frac{{3\pi }}{2}\)

Explanation:

Jacobian of (y₁, y₂, y₃) w.r.t. (x₁, x₂, x­₃) is equal to

\(\frac{{\partial \left( {{y_1},{y_2},\;{y_3}} \right)}}{{\partial \left( {{x_1},\;{x_2},\;{x_3}} \right)}} = \left| {\begin{array}{*{20}{c}}{\frac{{\partial {y_1}}}{{\partial {x_1}}}}&{\frac{{\partial {y_1}}}{{\partial {x_2}}}}&{\frac{{\partial {y_1}}}{{\partial {x_3}}}}\\{\frac{{\partial {y_2}}}{{\partial {x_1}}}}&{\frac{{\partial {y_2}}}{{\partial {x_2}}}}&{\frac{{\partial {y_2}}}{{\partial {x_3}}}}\\{\frac{{\partial {y_3}}}{{\partial {x_1}}}}&{\frac{{\partial {y_3}}}{{\partial {x_2}}}}&{\frac{{\partial {y_3}}}{{\partial {x_3}}}}\end{array}} \right|\)

\(= \left| {\begin{array}{*{20}{c}}{\frac{{ - {x_2} \cdot {x_3}}}{{x_1^2}}}&{\frac{{{x_3}}}{{{x_1}}}}&{\frac{{{x_2}}}{{{x_1}}}}\\{\frac{{{x_3}}}{{{x_2}}}}&{\frac{{ - {x_3} \cdot {x_1}}}{{x_2^2}}}&{\frac{{{x_1}}}{{{x_1}}}}\\{\frac{{{x_2}}}{{{x_3}}}}&{\frac{{{x_1}}}{{{x_3}}}}&{\frac{{ - {x_1} \cdot {x_2}}}{{x_3^2}}}\end{array}} \right|\)

\(= \frac{1}{{x_1^2 \cdot x_2^2 \cdot x_3^2}}\left| {\begin{array}{*{20}{c}}{ - {x_2}{x_3}}&{{x_1}{x_3}}&{{x_1}{x_2}}\\{{x_2}{x_3}}&{ - {x_1}{x_3}}&{{x_1}{x_2}}\\{{x_2}{x_3}}&{{x_1}{x_3}}&{ - {x_1}{x_2}}\end{array}} \right|\)

\(= \frac{{x_1^2x_2^2x_3^2}}{{x_1^2x_2^2x_3^2}}\left| {\begin{array}{*{20}{c}}{ - 1}&1&1\\1&{ - 1}&1\\1&1&{ - 1}\end{array}} \right|\)

= -1(1 - 1) - 1(-1 - 1) + 1(1 + 1)

= 0 + 2 + 2

= 4