Engineering Mathematics Test 3

\(\frac{{\partial f}}{{\partial y}} = \frac{\partial }{{\partial y}}\left( {{y^x}} \right) = x{y^{x - 1}}\) 

\(\frac{{{\partial ^2}f}}{{\partial x\partial y}} = \frac{\partial }{{\partial x}}\left( {x{y^{x - 1}}} \right) = x\frac{\partial }{{\partial x}}{y^{x - 1}} + {y^{x - 1}}\frac{\partial }{{\partial x}}\left( x \right)\;\) 

\(\therefore \frac{{{\partial ^2}f}}{{\partial x\partial y}} = x\;{y^{x - 1}}\log y + {y^{x - 1}}\) 

Putting x = 2 and y = 1,

The given region is, R: 0 ≤ x ≤ y ≤ L

The limits of x: 0 ≤ x ≤ L

The limits of y: x ≤ y ≤ L

\(\mathop \int\!\!\!\int \limits_R \left( {{x^2} + {y^2}} \right)dx\;dy\) 

\( = \mathop \smallint \limits_{x = 0}^{x = L} \mathop \smallint \limits_{y = x}^{y = L} \left( {{x^2} + {y^2}} \right)dx\;dy\) 

\(= \mathop \smallint \limits_{x = 0}^L \left[ {{x^2}y + \frac{{{y^3}}}{3}} \right]_x^2dx\) 

\( = \mathop \smallint \limits_{x = 0}^L \left[ {{x^2}L + \frac{{{L^3}}}{3} - \frac{{{x^3}}}{3} - {x^3}} \right]dx\;\) 

\(= \mathop \smallint \limits_{x = 0}^L \left[ {{x^2}L + \frac{{{L^3}}}{3} - \frac{{4{x^3}}}{3}} \right]dx\) 

\(= \left[ {L\frac{{{x^3}}}{3} + \frac{{{L^3}}}{3}x - \frac{{{x^4}}}{3}} \right]_0^L\) 

\(= \frac{{{L^4}}}{3} + \frac{{{L^4}}}{3} - \frac{{{L^4}}}{3} = \frac{{{L^4}}}{3}\) 

A and B are improper integrals since they have infinite (upper/lower) limits of integration.

D is an improper integral because the integrand 1 / x is undefined at \(x = 0\epsilon\left[ { - 2,\;2} \right]\).

C is not an improper integral because the integrand is defined everywhere in the finite limits of integration.

\(I = \mathop \smallint \limits_0^{\frac{\pi }{2}} \mathop \smallint \limits_0^{\frac{\pi }{2}} \mathop \smallint \limits_0^5 \sin \theta d\theta \;d\phi \;dr\)

\(I = \mathop \smallint \limits_0^{\frac{\pi }{2}} \sin \theta \;d\theta \mathop \smallint \limits_0^{\frac{\pi }{2}} d\phi \mathop \smallint \limits_0^5 dr\)

\( = \left[ { - \cos \theta } \right]_0^{\frac{\pi }{2}}\left[ \phi \right]_0^{\frac{\pi }{2}}\left[ r \right]_0^5\)

\(= 1 \times \frac{\pi }{2} \times 5 = 2.5\;\pi\)

We are given z = e^{ax + by}⋅ f(ax - by)

Now, \(\frac{{\partial z}}{{\partial x}} = a \cdot {e^{ax + by}} \cdot F\left( {ax - by} \right) + a{e^{ax + by}} \cdot F'\left( {ax - by} \right)\)

\(b\frac{{\partial z}}{{\partial x}} = ab\;[z + {e^{ax + by}} \cdot F'\left( {ax - by} \right)\) ---(1)

Now, \(\frac{{\partial z}}{{\partial y}} = b \cdot {e^{ax + by}} \cdot F\left( {ax - by} \right) - b \cdot {e^{ax + by}} \cdot F'\left( {ax - by} \right)\)

\(a\frac{{\partial z}}{{\partial y}} = ab\;\left[ {z - {e^{\left( {ax + by} \right)}} \cdot F'\left( {ax - by} \right)} \right]\) ---(2)

Add equation (1) & (2), we get

\(T = 2\pi \sqrt {\frac{l}{g}}\)

\(\log T = \log 2\pi + \frac{1}{2}\log l - \frac{1}{2}\log g\)

\(\frac{1}{T}\delta T = 0 + \frac{1}{2}\frac{{\delta l}}{l} - \frac{1}{2}\frac{{\delta g}}{g}\)

\(\frac{{\delta T}}{T}100 = \frac{1}{2}\left( {\frac{{\delta l}}{l}100 - \frac{{\delta g}}{g}100} \right)\)

\(= \frac{1}{2}\left( {1 \pm 2.5} \right)\)

= 1.75% or -0.75%