Category: Engineering Mathematics Objective Questions

250+ TOP MCQs on Partial Differentiation and Answers

Engineering Mathematics Multiple Choice Questions on “Partial Differentiation – 1”.

1. f(x, y) = x2 + xyz + z Find fx at (1,1,1)
a) 0
b) 1
c) 3
d) -1
Answer: c
Explanation: fx = 2x + yz
Put (x,y,z) = (1,1,1)
fx = 2 + 1 = 3.

2. f(x, y) = sin(xy) + x2 ln(y) Find fyx at (0, π⁄2)
a) 33
b) 0
c) 3
d) 1
Answer: d
Explanation: fy = xcos(xy) + x2⁄y
fyx = cos(xy) – xysin(xy) + 2x⁄y
Put (x,y) = (0, π⁄2)
= 1.

3. f(x, y) = x2 + y3 ; X = t2 + t3; y = t3 + t9 Find df⁄dt at t=1.
a) 0
b) 1
c)-1
d) 164
Answer: d
Explanation: Using chain rule we have
(frac{df}{dt}=f_x.frac{dx}{dt}+f_y.frac{dy}{dt})
=(2x).(2t + 3t2) + (3y2).(3t2 + 9t8)
Put t = 1; we have x = 2; y = 2
=4.(5) + 12.(12) = 164.

4. f(x, y) = sin(x) + cos(y) + xy2; x = cos(t); y = sin(t) Find df⁄dt at t = π⁄2
a) 2
b)-2
c) 1
d) 0
Answer: b
Explanation:Using chain rule we have
(frac{df}{dt}=f_x.frac{dx}{dt}+f_y.frac{dy}{dt})
= (cos(x) + y2).(-sin(t)) + (-sin(y) + 2xy).(cos(t))
Put t= π⁄2; we have x=0; y=1
=(1 + 1).(-1) + 0 = -2.

5. f(x, y, z, t) = xy + zt + x2 yzt; x = k3 ; y = k2; z = k; t = √k
Find df⁄dt at k = 1
a) 34
b) 16
c) 32
d) 61
Answer: b
Explanation: Using Chain rule we have
(frac{df}{dt}=f_x.frac{dx}{dk}+f_y.frac{dy}{dk}+f_z.frac{dz}{dk}+f_t.frac{dt}{dk})
= (y + 2xyzt).(3k2) + (x + x2zt).(2k) + (t + x2yt).(1) + (z + x2yz).((frac{1}{2sqrt{k}})
Put k=1; we have x=y=z=t=1
9 + 4 + 2 + 1 = 16.

6. The existence of first order partial derivatives implies continuity.
a) True
b) False
Answer: b
Explanation: The mere existence cannot be declared as a condition for contnuity because the second order derivatives should also be continuous.

7. The gradient of a function is parallel to the velocity vector of the level curve.
a) True
b) False
Answer: b
Explanation: The gradient is perpendicular and not parallel to the velocity vector of the level curve.

8. f(x, y) = sin(y + yx2) / 1 + x2 Value of fxy at (0,1) is
a) 0
b) 1
c) 67
d) 90
Answer: a
Explanation: First find
fy = cos(y + yx2)
Hence
fyx = fxy = – (2xy).sin(y + yx2)
Now put (x,y) = (0,1)
= 0.

9. f(x, y) = sin(xy + x3y) / x + x3 Find fxy at (0,1).
a) 2
b) 5
c) 1
d) undefined
Answer: c
Explanation: First find
fy = sin(xy + x3y)
Hence
fyx = fxy = (cos(xy + x3y)) . (y + 3x23y)
Now put (x,y) = (0,1)
= 1.

250+ TOP MCQs on Rectification in Polar and Parametric Forms and Answers

Differential and Integral Calculus Multiple Choice Questions on “Rectification in Polar and Parametric Forms”.

1. Find the length of the curve given by the equation.
(x^{frac{2}{3}}+y^frac{2}{3}=a^frac{2}{3})
a) (frac{3a}{2})
b) (frac{-7a}{2})
c) (frac{-3a}{4})
d) (frac{-3a}{2})
View Answer

Answer: d
Explanation: We know that,
S=(int_{x1}^{x2}sqrt{1+frac{dy}{dx}^2})
(y^frac{2}{3}=a^frac{2}{3}-x^frac{2}{3})
Differentiating on both sides
(frac{2}{3} y^{frac{2}{3}-1}= frac{-2}{3} x^{frac{2}{3}-1})
(frac{dy}{dx} = -frac{y}{x}^{frac{1}{3}})
((frac{dy}{dx})^2 = (frac{y}{x})^{frac{1}{3}})
(1+(frac{dy}{dx})^2=1+(frac{y}{x})^frac{2}{3})
Substituting from the original equation-
(1+(frac{dy}{dx})^2=(frac{a}{x})frac{2}{3})
(sqrt{1+frac{dy^2}{dx}}=(frac{a}{x})^{frac{1}{3}})
(S=int_{a}^{0}(frac{a}{x})^{frac{1}{3}} dx )
(s=frac{-3a}{2})
Thus, length of the given curve is (frac{-3a}{2}).

2. Find the length of one arc of the given cycloid.

x=a(θ-sinθ)
y=a(1+cosθ)

a) a
b) 4a
c) 8a
d) 2a
View Answer

Answer: c
Explanation: We know that
(s=int_{theta1}^{theta2}sqrt{(frac{dx}{dtheta})^2+(frac{dy}{dtheta})^2})
(frac{dx}{dtheta}=a(1-costheta))
(frac{dy}{dtheta}=a(-sintheta))
((frac{dx}{dtheta})^2+(frac{dy}{dtheta})^2=a^2(1-costheta)^2+a^2 sin^2theta)
((frac{dx}{dtheta})^2+(frac{dy}{dtheta})^2=4a^2 sin^2frac{theta}{2})
(s=int_{0}^{2}pisqrt{4a^2 sin^2frac{theta}{2}} dtheta)
On solving the given integral, we get
s=8a
Thus length of one arc of the given cycloid is 8a.

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250+ TOP MCQs on Bernoulli Equations and Answers

Ordinary Differential Equations Multiple Choice Questions on “Bernoulli Equations”.

1. Solution of the differential equation (frac{dy}{dx} + frac{y}{x} = y^2) x is ________
a) (frac{1}{y} = -x + c)
b) (frac{1}{xy} = -x + c)
c) (frac{1}{(xy)^2} = -y + c)
d) (frac{1}{xy} = -x^2 + c)
Answer: b
Explanation: Given equation is of the form (frac{dy}{dx} + Py = Qy^n) …divide by y2
where P and Q are functions of x hence this is Bernoulli’s equation in (y, frac{1}{y^2} frac{dy}{dx} + frac{1}{yx} = x)
put (frac{1}{y} = t rightarrow frac{-1}{y^2} frac{dy}{dx} = frac{dt}{dx}) substituting we get – (frac{dt}{dx} + frac{t}{x} = x ,or, frac{dt}{dx} – frac{t}{x} = -x)
this equation is linear in t i.e it is of the form
(frac{dt}{dx} + Pt = Q, I.F = e^{int P ,dx} =e^{int frac{-1}{x} ,dx} = e^{-log⁡x} = frac{1}{x})
its solution is
(te^{int P ,dx} = int Q e^{int P ,dx} ,dx + c rightarrow t frac{1}{x} = int -x * frac{1}{x} ,dx + c rightarrow frac{t}{x} = -x + c ,but, t = frac{1}{y})
thus solution is given by (frac{1}{xy} = -x + c).

2. Solution of the differential equation (frac{dy}{dx} -y ,tan⁡x = frac{sin⁡x cos^2⁡x}{y^2}) is ______
a) (y ,cos^2 x = frac{-cos^4 xsin^2 x}{2} + c)
b) (y^2 cos^2 x = frac{sin^6 x}{2} + c )
c) (y^3 cos^3 x = frac{-cos^6 x}{2} + c )
d) (y^4 cos^5 x = frac{sinx cos⁡x}{2} + c )
Answer: c
Explanation: (frac{dy}{dx} -y ,tan⁡x = frac{sin⁡x cos^2⁡x}{y^2}) multiplying by
(y^2 rightarrow y^2 frac{dy}{dx} – y^3 tan⁡x = sin⁡x ,cos^2⁡x )
put (t = y^3rightarrow3y^2 frac{dy}{dx} = frac{dt}{dx} ,or, y^2 frac{dy}{dx} = frac{1}{3} frac{dt}{dx})
substituting (frac{dt}{dx}-3t ,tan⁡x = 3sin⁡x ,cos^2⁡x )
this equation is linear in t i.e it is of the form (frac{dt}{dx} + Pt = Q, e^{int P ,dx} = e^{int -3tan⁡x ,dx})
(e^{-3 log⁡, sec⁡x} = cos^3 ,x) its solution is (te^{int P ,dx} = int Q ,e^{int P ,dx} dx + c)
t cos3 x=∫3sin⁡x cos2x cos3 x dx + c = ∫3sin⁡x cos5⁡x dx+c, put v=cos x
dv=-sin x dx i.e (int 3sin⁡x ,cos^5, ⁡x ,dx = int -3v^5 ,dx = frac{-v^6}{2} = frac{-cos^6 x}{2}) hence its solution becomes (t cos^3 x = frac{-cos^6 x}{2} + c rightarrow y^3 cos^3 ,x = frac{-cos^6 x}{2} + c.)

3. Solution of the differential equation 6y2 dx – x(x3 + 2y)dy = 0 is ________
a) (frac{y}{x^3} = frac{-log⁡y}{2} + c )
b) (frac{y^2}{x^3} = frac{-log⁡x}{4} + c )
c) (frac{x}{y^3} = frac{-log⁡x}{2} + c )
d) (frac{x}{y^2} = frac{-log⁡y}{4} + c )
Answer: a
Explanation: Equation is reduced to (frac{dy}{dx} = frac{x(x^3+2y)}{6y^2} ,i.e, frac{dx}{dy}-frac{x}{3y} = frac{x^4}{6y^2}) …divide by x4
we get (frac{1}{x^4} frac{dy}{dx} – frac{1}{3x^3 y} = frac{1}{6y^2} ,put, frac{1}{x^3} = t rightarrow frac{-3}{x^4} frac{dx}{dy} = frac{dt}{dy} ,or, frac{1}{x^4} frac{dy}{dx} = frac{-1}{3} frac{dt}{dy} )
substituting (frac{-1}{3} frac{dt}{dy} – frac{t}{3y} = frac{1}{6y^2} ,or, frac{dt}{dy} + frac{t}{y} = frac{-1}{2y^2}) this equation is linear in t i.e it is of the form
(frac{dt}{dy} + Pt = Q ,where, P=frac{1}{y}, Q=frac{-1}{2y^2} ,I.F, = e^{int P ,dy} = e^{int frac{1}{y} ,dy} = e^{log⁡y} = y) its solution is (te^{int P ,dy} = int Q ,e^{int P ,dy} + c rightarrow ty = int y * frac{-1}{2y^2} ,dy + c = frac{-log⁡y}{2} + c)
(frac{y}{x^3} = frac{-log⁡y}{2} + c ) is the required solution.

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250+ TOP MCQs on Laplace Transform by Properties and Answers

Engineering Mathematics Multiple Choice Questions on “Laplace Transform by Properties – 1”.

1. Laplace of function f(t) is given by?
a) F(s)=(int_{-infty}^infty f(t)e^{-st} ,dt)
b) F(t)=(int_{-infty}^infty f(t)e^{-t} ,dt)
c) f(s)=(int_{-infty}^infty f(t)e^{-st} ,dt)
d) f(t)=(int_{-infty}^infty f(t)e^{-t} ,dt)
Answer: a
Explanation: Laplace of function f(t) is given by
F(s)=(int_{-infty}^infty f(t)e^{-st} ,dt).

2. Laplace transform any function changes it domain to s-domain.
a) True
b) False
Answer: a
Explanation: Laplace of function f(t) is given by F(s)=(int_{-infty}^infty f(t)e^{-st} ), hence it changes domain of function from one domain to s-domain.

3. Laplace transform if sin⁡(at)u(t) is?
a) s ⁄ a2+s2
b) a ⁄ a2+s2
c) s2 ⁄ a2+s2
d) a2 ⁄ a2+s2
Answer: b
Explanation: We know that,
F(s)=(int_{-infty}^infty sin⁡(at)u(t) e^{-st} dt=int_0^∞ sin⁡(at)e^{-st} dt)
=(left [frac{e^{-st}}{a^2+s^2}[-ssin(at)-acos⁡(at)]right ]_∞^0)
=(frac{a}{a^2+s^2})

4. Laplace transform if cos⁡(at)u(t) is?
a) s ⁄ a2+s2
b) a ⁄ a2+s2
c) s2 ⁄ a2+s2
d) a2 ⁄ a2+s2
Answer: a
Explanation: We know that,
F(s)=(int_{-infty}^infty cos(at)u(t) e^{-st} dt=int_0^∞ cos⁡(at)e^{-st} dt)
=(left [frac{e^{-st}}{a^2+s^2}[-scos(at)-asin⁡(at)]right ]_∞^0)
=(frac{a}{a^2+s^2})

5. Find the laplace transform of et Sin(t).
a) (frac{a}{a^2+(s+1)^2})
b) (frac{a}{a^2+(s-1)^2})
c) (frac{s+1}{a^2+(s+1)^2})
d) (frac{s+1}{a^2+(s+1)^2})
Answer: b
Explanation:
F(s)=(int_{-infty}^infty e^t sin⁡(at)u(t) e^{-st} dt=∫_0^∞ sin⁡(at)e^{-(s-1)t} dt)
=(left [frac e^{-st}{a^2+(s-1)^2} [-(s-1)sin(at)-acos⁡(at) ]right ]_0^∞)
=(frac{a}{a^2+(s-1)^2})

6. Laplace transform of t2 sin⁡(2t).
a) (left [frac{12s^2-16}{(s^2+4)^4}right ])
b) (left [frac{3s^2-4}{(s^2+4)^3}right ])
c) (left [frac{12s^2-16}{(s^2+4)^6}right ])
d) (left [frac{12s^2-16}{(s^2+4)^3}right ])
Answer: d
Explanation: We know that,
(L(t^n f(t))=(-1)^n frac{d^n F(s)}{ds^n}),
Here, f(t)=sin⁡(2t)=>F(s)=(frac{2}{s^2+4}),
Hence, (L(t^2 sin⁡(2t))=frac{d^2}{ds^2} (frac{2}{s^2+4})=frac{d}{ds} frac{(s^2+4).0-2(2s)}{(s^2+4)^2})
=(-4left [frac{(s^2+4)^2-2s(s^2+4)2s}{(s^2+4)^4} right ]=left [frac{12s^2-16}{(s^2+4)^3}right ])

7. Find the laplace transform of t5⁄2.
a) (frac{15}{8} frac{√π}{s^{5/2}})
b) (frac{15}{8} frac{√π}{s^{7/2}})
c) (frac{9}{4} frac{√π}{s^{7/2}})
d) (frac{15}{4} frac{√π}{s^{7/2}})
Answer: b
Explanation:
(g(t)=t^{5/2}=frac{5}{2} int_0^t t^{frac{3}{2}} dt=frac{15}{4} int_0^t int_0^t √t dt dt)
let f(t)=√t, hence, F(s)=(frac{sqrt{π}}{2s^{frac{3}{2}}})
hence, G(s)=(frac{15}{4} ,frac{1}{s^2} ,F(s)=frac{15}{8} frac{√π}{s^{7/2}})

8. Value of (int_{-infty}^infty e^t ,Sin(t)Cos(t)dt) = ?
a) 0.5
b) 0.75
c) 0.2
d) 0.71
Answer: c
Explanation: L(Sin(2t)) = (int_{-infty}^infty e^{-st} ,Sin(2t)dt) = 2/(s2 + 4)
Putting s=-1
(int_{-infty}^infty e^t ,Sin(2t)dt) = 0.4
hence,
(int_{-infty}^infty e^{-st} ,Sin(t)Cos(t)dt) = 0.2.

9. Value of (int_{-infty}^infty e^t ,Sin(t) ,dt) = ?
a) 0.50
b) 0.25
c) 0.17
d) 0.12
Answer: a
Explanation: L(Sin(2t)) = (int_{-infty}^infty e^{-st} ,Sin(t)dt) = 1/(s2 + 1)
Putting s = -1
(int_{-infty}^infty e^t ,Sin(t)dt) = 0.5.

10. Value of (int_{-infty}^infty e^t ,log(1+t)dt) = ?
a) Sum of infinite integers
b) Sum of infinite factorials
c) Sum of squares of Integers
d) Sum of square of factorials
Answer: b
Explanation:
(int_{-infty}^infty e^t (t-t^2/2+t^3/3-….)dt)
(int_{-infty}^infty te^t dt=0.5 int_{-infty}^infty te^t dt)
Now,
(int_{-infty}^infty te^t dt– 1/2 int_{-infty}^infty t^2 e^t dt + (1/3) int_{-infty}^infty t^3 e^t dt-………)
Now, (int_{-infty}^infty t^n e^t dt=n!/(-1)^{n+1})
Hence,
(int_{-infty}^infty t^n e^t dt = 1 – (1/2)(2!/(-1)^3) + (1/3)(3!/)-…….)
(int_{-infty}^infty t^n e^t dt) = 0! + 1! + 2! + 3! +…. = Sum of infinite factorials.

11. Find the laplace transform of y(t)=et.t.Sin(t)Cos(t).
a) (frac{4(s-1)}{[(s-1)^2+4]^2})
b) (frac{2(s+1)}{[(s+1)^2+4]^2})
c) (frac{4(s+1)}{[(s+1)^2+4]^2})
d) (frac{2(s-1)}{[(s-1)^2+4]^2})
Answer: d
Explanation:
y(t)=(frac{1}{2} t.e^t Sin(2t))
Laplace transform of Sin(2t)=(frac{2}{s^2+4})
Laplace transform of tSin(2t)=(-frac{d}{dt} frac{2}{s^2+4}=frac{2(2s)}{(s^2+4)^2}=frac{4s}{(s^2+4)^2})
Laplace transform of te^t Sin(2t)=(frac{4(s-1)}{[(s-1)^2+4]^2})
Laplace transform of 1/2 tet Sin(2t)=(frac{2(s-1)}{[(s-1)^2+4]^2})

12. Find the value of (int_0^{infty} tsin(t)cos(t)).
a) s ⁄ s2+22
b) a ⁄ a2+s4
c) 1
d) 0
Answer: d
Explanation:
y(t)=(frac{1}{2} t Sin(2t)u(t))
Laplace transform of Sin(2t)=(frac{2}{s^2+4})
Laplace transform of tSin(2t)=(-frac{d}{dt} frac{2}{s^2+4}=frac{2(2s)}{(s^2+4)^2}=frac{4s}{(s^2+4)^2})
Laplace transform of (frac{1}{2}tsin(2t)=int_{-0}^{infty} e^{-st} tsin(t)cos(t)dt=frac{2s}{[s^2+4]^2})
Putting, s = 0, (int_0^{infty} tsin(t)cos(t)dt=0)

13. Find the laplace transform of y(t)=e|t-1| u(t).
a) (frac{2s}{1-s^2} e^s)
b) (frac{2s}{1+s^2} e^{-s})
c) (frac{2s}{1+s^2} e^s)
d) (frac{2s}{1-s^2} e^{-s})
Answer: d
Explanation:
y(t)=(e^{|t-1|})
Laplace transform of e|t| =(int_{-infty}^infty e^{|t|} e^{-st} dt)
=(int_0^∞ e^t e^{-st} dt-int_{-∞}^0 e^{-t} e^{-st} dt)
=(int_0^∞ e^{-(s-1)t} dt-∫_{-∞}^0 e^{(-s-1)t} dt)
Now,(int_0^∞ e^{-(s-1)t} dt=left [-frac{1}{s-1} [e^{-(s-1)t}]right ]_∞^0)
=(left [-frac{1}{s-1} [e^{-(s-1)t} ]right ]_∞^0=frac{-1}{s-1})
Now, (∫_{-∞}^0 e^{(-s-1)t} dt=left [frac{1}{-(s+1)} [e^{(-s-1)t}]right ]_0^{-∞})
=(left [-frac{1}{s+1} [e^{(-s-1)t} ]right ]_0^{-∞}=-frac{1}{(s+1)})
Laplace transform of |t| e|t| =(int_{-infty}^infty e^{|t|} e^{-st} dt=-left [frac{1}{s-1}+frac{1}{s+1}right ]=-left [frac{2s}{s^2-1}right ])
Laplace transform of |t| e|t| = (int_{-infty}^infty e^{|t-1|} e^{-st} dt=frac{2s}{1-s^2} e^{-s})

250+ TOP MCQs on Derogatory and Non-Derogatory Matrices and Answers

Matrices Multiple Choice Questions on “Derogatory and Non-Derogatory Matrices”.

1. Identify the type of Matrix
(begin{bmatrix}7&4&-1\4&7&-1\-4&-4&4end{bmatrix})
a) Identity Matrix
b) Non Derogatory Matrix
c) Derogatory Matrix
d) Symmetrical Matrix
Answer: c
Explanation: In this we have,

A=(begin{bmatrix}7&4&-1\4&7&-1\-4&-4&4end{bmatrix})

Finding the Characteristic Equation of the matrix
γ3-18γ2+81γ-108=0
Solving the Characteristic equation, we get the Eigenvalues
γ=3,3,12
Since Eigenvalues are repeated, this is a Derogatory Matrix.

2. Find the Eigenvalues and the type of the given matrix.
A=(begin{bmatrix}3&10&5\-2&-3&-4\3&5&7end{bmatrix})
a) 3, 1, 3 Non Derogatory
b) 2, 2, 2 Derogatory
c) 3, 2, 2 Derogatory
d) 1, 2, 3 Non Derogatory
Answer: c
Explanation: In this we have,
A=(begin{bmatrix}3&10&5\-2&-3&-4\3&5&7end{bmatrix})
Finding the characteristic equation of the Matrix,
-γ3+3γ2+24γ-141=0
Solving the Characteristic equation, we get the Eigenvalues
γ=3, 2, 2
Since Eigenvalues are repeated, this is a Derogatory Matrix.

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250+ TOP MCQs on First Order PDE and Answers

Partial Differential Equations Multiple Choice Questions on “First Order PDE”.

1. Find (frac{partial z}{partial x}) where (z=ax^2+2by^2+2bxy).
a) 3by
b) 2ax
c) 3(ax+by)
d) 2(ax+by)
View Answer

Answer: d
Explanation: Here we use partial differentiation.
(z=ax^2+2by^2+2bxy).
(frac{partial z}{partial x})=a(2x)+0+2by
(frac{partial z}{partial x})=2ax+2by
(frac{partial z}{partial x})=2(ax+by).

2. Find (frac{partial z}{partial x}) where (z=sin⁡x^2×cos⁡y^2).
a) 2xsin⁡x2
b) x sin2x
c) 2xsin⁡x2 cos⁡y2
d) 6xsin⁡x2 cos⁡y2
View Answer

Answer: c
Explanation: Here we use partial differentiation.
z=sin⁡x2×cos⁡y2
(frac{partial z}{partial x}=(cos⁡x^2×2x)× cos⁡y^2)
(frac{partial z}{partial x}=2xsin⁡x^2 cos⁡y^2).

3. Find (frac{partial u}{partial x}) where (u=cos⁡(sqrt x+sqrt y)).
a) (frac{-1}{2sqrt x}×tan⁡(sqrt x+sqrt y))
b) (frac{-1}{2sqrt x}×cos⁡(sqrt x+sqrt y))
c) (frac{-1}{2sqrt x}×sin⁡(sqrt x+sqrt y))
d) (frac{-1}{sqrt x}×sin⁡(sqrt x+sqrt y))
View Answer

Answer: c
Explanation: Here we use partial differentiation.
(u=cos⁡(sqrt x+sqrt y))
(frac{partial u}{partial x}= -sin⁡(sqrt x+sqrt y)×frac{1}{2sqrt x})
(frac{partial z}{partial x}=frac{-1}{2sqrt x}×sin⁡(sqrt x+sqrt y)).

4. If (u=frac{e^{x+y}}{e^x-e^y}), what is (frac{partial u}{partial x}+frac{partial u}{partial y})?
a) (frac{2((e^x-e^y)×e^{x+y})-(e^{x+y}) (e^x+e^y)}{(e^x-e^y)^2} )
b) (frac{2((e^x-e^y)×e^{x+y})-(e^{x+y}) (e^x+e^y)}{(e^x+e^y)^2} )
c) (frac{2((e^x-e^y)×e^{x+y})-(e^{x+y}) (e^x-e^y)}{(e^x-e^y)^2} )
d) u
View Answer

Answer: a
Explanation: Here we use partial differentiation.
(frac{partial u}{partial x}=frac{(e^x-e^y)×e^{x+y}-(e^{x+y})(e^x)}{(e^x-e^y)^2})
(frac{partial u}{partial y}=frac{(e^x-e^y)×e^{x+y}-(e^{x+y})(e^y)}{(e^x-e^y)^2})
(frac{partial u}{partial x}+frac{partial u}{partial y}=frac{(e^x-e^y)×e^{x+y}-(e^{x+y})(e^x)}{(e^x-e^y)^2} + frac{(e^x-e^y)×e^{x+y}-(e^(x+y))(e^y)}{(e^x-e^y)^2} )
(frac{partial u}{partial x}+frac{partial u}{partial y}=frac{2((e^x-e^y)×e^{x+y})-(e^{x+y}) (e^x+e^y)}{(e^x-e^y)^2}).

5. If (theta=t^n e^frac{-r^2}{2t}), find the value of n that satisfies the equation, (frac{partial theta}{partial t}=frac{1}{r^2}frac{partial}{partial r}(r^2 frac{partial theta}{partial r})).
a) 0
b) -1
c) 1
d) 3
View Answer

Answer: b
Explanation: Here we use partial differentiation.
(frac{partial theta}{partial t}=nt^{n-1}×e^frac{-r^2}{2t}+t^n×e^frac{-r^2}{2t}×frac{r^2}{2t^2})
(frac{partial theta}{partial t}=nt^{n-1}×e^frac{-r^2}{2t}+t^n×e^frac{-r^2}{2t}×frac{r^2}{2t^2})
Substituting from the question
(frac{partial theta}{partial t}=frac{ntheta}{t}+frac{r^2 theta}{2t^2})
(frac{partial theta}{partial t}=(frac{n}{t}+frac{r^2}{2t^2})theta )
(frac{partial theta}{partial r}=t^n×e^frac{-r^2}{2t}×frac{-r}{t})
Substituting from the question
(frac{partial theta}{partial r}=frac{-rtheta}{t})
(r^2 frac{partial theta}{partial r}=frac{-r^3 theta}{t})
Now substituting the values of (frac{partial theta}{partial r}) and (frac{partial theta}{partial t}) in the original equation,
(left ( frac{n}{t}+frac{r^2}{2t^2} right )theta = frac{1}{r^2} frac{partial}{partial r}(frac{-r^3 theta}{t}))
(frac{n}{t}=frac{-1}{t})
n=-1.

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