300+ REAL TIME Engineering Mathematics Objective Questions & Answers 2023

250+ TOP MCQs on Taylor’s Theorem Two Variables and Answers

Differential and Integral Calculus Multiple Choice Questions on “Taylor’s Theorem Two Variables”.

1. Among the following which is the correct expression for Taylor’s theorem in two variables for the function f (x, y) near (a, b) where h=x-a & k=y-b upto second degree?
a) (f (a+ h, b+ k) = f (a, b) + frac{x-a}{1!} f_x(a, b) + frac{y-b}{1!} f_y(a, b) + frac{(x-a)^2}{2!} f_{xx}(a, b)\
+2 frac{(x-a)(y-b)}{4!} f_{xy} (a,b) + frac{(y-b)^2}{2!} f_{yy}(a, b))
b) (f (a+ h, b+ k) = f (a, b) + frac{x-a}{1!} f_x(a, b) + frac{y-b}{1!} f_y(a, b) + frac{(x-a)^2}{2!} f_{xx}(a, b)\
+ frac{(x-a)(y-b)}{2!} f_{xy} (a,b) + frac{(y-b)^2}{2!} f_{yy}(a, b))
c) (f (a+ h, b+ k) = f (a, b) + frac{x-a}{1!} f_x(a, b) + frac{y-b}{1!} f_y(a, b) + frac{(x-a)^2}{2!} f_{xx}(a, b)\
+ frac{(y-b)^2}{2!} f_{yy}(a, b))
d) (f (a+ h, b+ k) = f (a, b) + frac{x-a}{1!} f_x(a, b) + frac{y-b}{1!} f_y(a, b) + frac{(x-a)^2}{2!} f_{xx}(a, b)\
+2 frac{(x-a)(y-b)}{2!} f_{xy} (a,b) + frac{(y-b)^2}{2!} f_{yy}(a, b))
Answer: d
Explanation: By definition
(f (a+ h, b+ k) = f (a, b) + frac{x-a}{1!} f_x(a, b) + frac{y-b}{1!} f_y(a, b) + frac{(x-a)^2}{2!} f_{xx}(a, b)\
+2 frac{(x-a)(y-b)}{2!} f_{xy} (a,b) + frac{(y-b)^2}{2!} f_{yy}(a, b))
here we can observe that second degree is of the form (p+q)² similarly Taylor’s theorem is expanded to third degree which is of the form (p+q)³ & f (a+ h, b+ k) = f (x, y)
where((f_x=frac{∂f (x,y)}{∂x}, f_y=frac{∂f (x,y)}{∂y}, f_{xx}=frac{∂}{∂x}(frac{∂f(x,y)}{∂x}), f_{yy}=frac{∂}{∂y} (frac{∂f (x,y)}{∂y}), \
f_{xy}=frac{∂}{∂x}(frac{∂f (x,y)}{∂y})).)

2. Given f (x,y)=e^x cos⁡y, what is the value of the fifth term in Taylor’s series near (1,(frac{π}{4})) where it is expanded in increasing order of degree & by following algebraic identity rule?
a) (frac{-e(x-1)(y-frac{π}{4})}{sqrt{2}})
b) (-sqrt{2} e(x-1)(y-frac{π}{4}))
c) (frac{e(x-1)^2}{sqrt{2}})
d) (frac{e(y-frac{π}{4})^2}{sqrt{2}})
Answer: a
Explanation: Taylor’s series expansion is given by
(f (a+ h, b+ k) = f (a, b) + frac{x-a}{1!} f_x(a, b) + frac{y-b}{1!} f_y(a, b) + frac{(x-a)^2}{2!} f_{xx}(a, b)\
+2 frac{(x-a)(y-b)}{2!} f_{xy} (a,b) + frac{(y-b)^2}{2!} f_{yy}(a, b))
Thus fifth term is given by (2 frac{(x-a)(x-b)}{2!} f_{xy} (a,b))..(1) where a=1, b=π/4 & (f_{xy}=frac{∂}{∂x}(frac{∂f(x,y)}{∂x}) = frac{∂}{∂x}(frac{∂e^x cos⁡y}{∂y}) =- e^x ,sin⁡y ) at (1, (frac{π}{4}), f_{xy}=frac{-e}{sqrt{2}}) substituting in (1)
We get fifth term as (2 frac{(x-1)(x-π/4)}{2!} frac{-e}{sqrt{2}} = frac{-e(x-1)(y-frac{π}{4})}{sqrt{2}}).

3. Given f (x,y)=sin⁡xy, what is the value of the third degree first term in Taylor’s series near (1,-(frac{π}{2})) where it is expanded in increasing order of degree & by following algebraic identity rule?
a) (frac{π^3}{8})
b) (frac{π^3}{8} frac{(x-1)(y+frac{π}{2})}{3!})
c) 0
d) (-frac{π^3}{8} frac{(x-1)^3}{3!})
Answer: c
Explanation: Third degree first term in Taylor’s series is given by (frac{(x-a)^3 f_{xxx} (x,y)}{3!}) Where a=1 (b=-frac{π}{2}, f_{xxx} (x,y)=frac{∂^3 f(x,y)}{∂x^3} ,i.e, frac{∂^3 sin⁡xy}{∂x^3} = -y^3 cos⁡xy)…… (partial differentiating f (x,y) w.r.t x only)
at (a=1, b=-frac{π}{2}, frac{∂^3 sin⁡xy}{∂x^3} = -frac{π^3 cos-frac{⁡π}{2}}{8}=0) hence third degree first term is given by (-frac{π^3}{8} frac{(x-1)^3}{3!}.0 = 0.)

4. Taylor’s theorem is mainly used in expressing the function as sum with infinite terms.
a) True
b) False
Answer: a
Explanation: Taylor’s theorem helps in expanding a function into infinite terms however, it can be applied to functions that can be expressed finitely.

5. Expansion of (f (x,y) = tan^{-1} frac{⁡y}{x}) upto first degree containing (x+1) & (y-1) is __________
a) (frac{3π}{4} + frac{(x+1)}{1!} frac{-1}{2} + frac{(y-1)}{1!} frac{-1}{2} + frac{(x+1)^2}{2!} frac{-1}{2} + frac{(y-1)^2}{2!} frac{1}{2})
b) (frac{π}{4} + frac{(x+1)}{1!} frac{-1}{2} + frac{(y-1)}{1!} frac{-1}{2} + frac{(x+1)^2}{2!} frac{1}{4} + frac{(y-1)^2}{2!} frac{1}{4})
c) (frac{5π}{4} + frac{(x+1)}{1!} frac{-1}{2} + frac{(y-1)}{1!} frac{-1}{2} + frac{(x+1)^2}{2!} frac{-1}{4} + frac{(y-1)^2}{2!} frac{1}{4})
d) (frac{3π}{4} + frac{(x+1)}{1!} frac{-1}{2} + frac{(y-1)}{1!} frac{-1}{2} + frac{(x+1)^2}{2!} frac{-1}{4} + frac{(y-1)^2}{2!} frac{1}{4})
Answer: a
Explanation: We can expand the given function according to Taylor’s theorem
(f (a+ h, b+ k) = f (a, b) + frac{x-a}{1!} f_x(a, b) + frac{y-b}{1!} f_y(a, b) + frac{(x-a)^2}{2!} f_{xx}(a, b)\
+2 frac{(x-a)(y-b)}{2!} f_{xy} (a,b) + frac{(y-b)^2}{2!} f_{yy}(a, b))
Given a=-1 & b=1, f(-1,1)=tan^-1⁡-1 = (frac{3π}{4})
(f_x = frac{-y}{x^2+y^2} ,at, (-1,1) = frac{-1}{2})
(f_y = frac{x}{x^2+y^2} ,at, (-1,1) = frac{-1}{2})
(f_{xy} = frac{(x^2+y^2)-2x^2}{(x^2+y^2)^2}) at (-1,1)=0
(f_{xx} = frac{2yx}{(x^2+y^2)^2} ,at, (-1,1)=frac{-2}{4} = frac{-1}{2})
(f_{yy} = frac{-2yx}{(x^2+y^2)^2} ,at, (-1,1)=frac{2}{4} = frac{1}{2}) thus the series is given by
(frac{3π}{4} + frac{(x+1)}{1!} frac{-1}{2} + frac{(y-1)}{1!} frac{-1}{2} + frac{(x+1)^2}{2!} frac{-1}{2} + frac{(y-1)^2}{2!} frac{1}{2}).

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250+ TOP MCQs on Change of Variables In a Double Integral and Answers

Differential and Integral Calculus Questions and Answers for Experienced people focuses on “Change of Variables In a Double Integral”.

1. Evaluation of (intint_R f(x,y) ,dx ,dy ) in cartesian coordinate can be done using change of variables principle, among the choices given below which is correct explanation of change of variables principle? (Given let x=g(u,v) & y=h(u,v))
a) (intint_S f(g(u,v),h(u,v)) ,du ,dv)
b) (intint_S f(g(u,v),h(u,v)) frac{d(x,y)}{d(u,v)} ,du ,dv)
c) (intint_S f(g(u,v),h(u,v)) frac{∂(x,y)}{∂(u,v)} ,du ,dv)
d) (intint_S f(g(u,v),h(u,v)) frac{∂(u,v)}{∂(x,y)} ,du ,dv)
Answer: c
Explanation: (intint_S f(g(u,v),h(u,v)) frac{∂(x,y)}{∂(u,v)} ,du ,dv)
where ( frac{∂(x,y)}{∂(u,v)} = begin{vmatrix}
frac{∂x}{∂u} frac{∂x}{∂v}\
frac{∂y}{∂u} frac{∂y}{∂v}\
end{vmatrix} = J(frac{x,y}{u,v}) = frac{∂x}{∂u} frac{∂y}{∂v} – frac{∂x}{∂v} frac{∂y}{∂u} …..)(‘J’ is Jacobian).

2. The value of ∬_R (x-y)² dx dy where R is the parallelogram with vertices (0,0), (1,1),(2,0), (1,-1) when solved using change of variables is given by____

a) 16/3
b) 8/3
c) 4/3
d) 0
Answer: b
Explanation: W.K.T from change of variables principle
(intint_R f(x,y),dx ,dy = intint_S f(g(u,v),h(u,v)) frac{∂(x,y)}{∂(u,v)} ,du ,dv …..(1))
From the above diagram in the region R the equations are given by
x-y=0, x-y=2, x+y=0, x+y=2 from this we can observe that change of
variables is u=x-y, v=x+y solving we get (x=frac{u+v}{2}, y=frac{v-u}{2})
(frac{∂(x,y)}{∂(u,v)} = begin{vmatrix}
frac{∂x}{∂u}&frac{∂x}{∂v}\
frac{∂y}{∂u}&frac{∂y}{∂v}\
end{vmatrix} = begin{vmatrix}
0.5 &0.5\
-0.5&0.5\
end{vmatrix} = 0.5)
The region S in the (u,v) is the square 0becomes (int_0^2 int_0^2 0.5u^2 ,du ,dv)……from(1)
(=int_0^2 Big[frac{u^3}{6}Big]_0^2 ,dv = int_0^2 frac{4}{3} dv = frac{8}{3}.)

3. If double integral in Cartesian coordinate is given by ∬_R f(x,y) dx dy then the value of same integral in polar form is _____
a) ∬_P f(r cos θ, rsin θ)dr dθ
b) ∬_P f(r cos⁡θ, r sin⁡θ)rdr dθ
c) ∬_P f(r cos⁡θ, r sin⁡θ) r² dr dθ
d) ∬_P f(r sin⁡θ, r cos⁡θ)dr dθ
Answer: b
Explanation: ∬_R f(x,y)dx dy when converting this into polar form we take x = r cos θ
y=r sin θ as change of variables from
(intint_R f(x,y) ,dx ,dy = intint_S f(g(u,v),h(u,v)) frac{∂(x,y)}{∂(u,v)} ,du ,dv ) where u=r & v=θ
thus (frac{∂(x,y)}{∂(r,θ)} =begin{vmatrix}
frac{∂x}{∂r} &frac{∂x}{∂θ}\
frac{∂y}{∂r} &frac{∂y}{∂θ}\
end{vmatrix} = begin{vmatrix}
cos⁡θ &-r sin⁡θ\
sin⁡θ & rcos⁡θ\
end{vmatrix} = r(cos^2 θ + sin^2 θ) = r)
substituting we get ∬_P f(r cos θ,r sin θ)rdr dθ.

4. The value of ∬_R sin⁡(x² + y²) dx dy where R is the region bounded by circle centered at origin with radius r=2 is _____
a) πcos 4
b) π(1-cos 4)
c) π
d) π(1-sin 4)
Answer: b
Explanation: Using Polar variable transformation x = r cos θ & y=r sin θ, r varies from 0 to 2 & θ varies from 0 to 2π because radius of circle i.e r=2 & centered at origin
(intint_P f(r cos θ,r sin θ )rdr ,dθ = intint_R sin⁡ (x^2+y^2) ,dx ,dy = int_0^2π int_0^2 r sin⁡ r^2 ,dr ,dθ)
Using substitution t=r² integral changes to (int_0^{2π} int_0^4 0.5 ,sin⁡t ,dt ,dθ)
(int_0^{2π} 0.5big[-cos⁡tbig]_0^4 dθ = int_0^{2π} 0.5(1-cos⁡4) dθ = π(1-cos 4)).

5. Using change of variables principle in double integral we can reduce cartesian integral to simpler form.
a) True
b) False
Answer: a
Explanation: The above statement is not necessarily true always but change of variables cartesian to polar form & changing into suitable form reduces the complexity in the evaluation of double integral however converse is also true for example ∬_R x² y³ dx dy is the given integral while solving there is no need of converting it into polar form since ∬_P r⁶ cos² θ sin³ θ dr dθ is tedious to solve when R only consists of constants but ∬_R x² y³ dx dy can be solved easily by ordinary method of integration.

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250+ TOP MCQs on Newton’s Law of Cooling and Escape Velocity and Answers

Ordinary Differential Equations Multiple Choice Questions on “Newton’s Law of Cooling and Escape Velocity”.

1. According to Newton’s law of cooling “The change of temperature of a body is proportional to the difference between the temperature of a body and that of the surrounding medium”. If t₁℃ is the initial temperature of the body and t₂℃ is the constant temperature of the medium, T℃ be the temperature of the body at any time t then find the expression for T℃ as a function of t₁℃, t₂℃ and time t.
a) T=t₁+(t₂) e^-kt
b) T=t₂+(t₁-t₂) e^-kt
c) T=t₁+(t₁-t₂) e^kt
d) T=t₂+(t₁) e^kt
Answer: b
Explanation: According to the definition of Newton’s law of cooling (frac{dT}{dt} ∝ (T-t_2) ,or, frac{dT}{dt} = -k(T-t_2)) ….k is a constant of proportionality and negative sign indicates the cooling of a body with increase of the time. since t₁℃ initial temperature of the body at t=0 T=t₁ –> T(0) = t₁℃. (frac{dT}{dt} = -k(T-t_2))…….at T(0) = t₁℃, now solving DE the above equation is of variable separable form i.e (int frac{dt}{T-t_2} = int -kdt + c)
=log (T-t₂) = -kt + c –> T-t₂ = pe^-kt…where p=e^c=constant, using initial condition i.e T(0)= t₁ we get t₁-t₂=p substituting back in equation we obtain T=t₂+(t₁-t₂) e^-kt.

2. A body in air at 25℃ cools from 100℃ to 75℃ in 1 minute. What is the temperature of the body at the end of 3 minutes? (Take log(1.5)=0.4)
a) 40℃
b) 47.5℃
c) 42.5℃
d) 50℃
Answer: b
Explanation: By Newton’s law of cooling w.k.t T = t₂ + (t₁-t₂) e^-kt, given t₁=100℃, t₂=25℃
when t=1 –> T(1) = 25 + 75 e^-k = 75℃ –> 50/75 = 2/3 = e^-k
–> 3/2=e^k taking log k=log(1.5)=0.4.
to find T when t=3 minute using the value of k we get
T = 25 + 75e^-0.4*3 = 47.5℃……e^-1.2=0.3.

3. A bottle of mineral water at a room temperature of 72℉ is kept in a refrigerator where the temperature is 44℉.After half an hour water cooled to 61℉.What is the temperature of the body in another half an hour?(Take log (frac{28}{17}) = 0.498, e^-0.99=0.37)
a) 18℉
b) 9.4℉
c) 54.4℉
d) 36.4℉
Answer: c
Explanation: By Newton’s law of cooling w.k.t T=t₂+(t₁-t₂) e^-kt, given t₁=72℉, t₂=44℉
At t=half an hour = 30mts T=61℉, finding k using the given values i.e
61=44+28e^-k30 –> (frac{17}{28}) = e^-k30 or (frac{28}{17}) = e^k30 taking log, log (frac{28}{17}) = 30k –> k=0.0166
to find T when t = 30mts + 30mts = 60mts
T = 44 + 28e^-(0.0166)30 = 54.4℉.

4. The radius of the moon is roughly 2000km. The acceleration of gravity at the surface of the moon is about (frac{g}{6}), where g is the acceleration of gravity at the surface of the earth. What is the velocity of escape for the moon?(Take g=10ms^-2)
a) 2.58 kms^-1
b) 4.58 kms^-1
c) 6.28 kms^-1
d) 12.28 kms^-1
Answer: a
Explanation: Let R be radius of the earth and r be the variable distance from Newton’s law (a = frac{dv}{dt} = frac{k}{r^2}) when r=R a=-g due to retardation of a body -gR²=k
substituting the value of k back we get (frac{dv}{dt} = frac{-gR^2}{r^2} = frac{dr}{dt} frac{dv}{dr} = v frac{dv}{dr}) solving DE for v
we get (int v ,dv = int frac{-gR^2}{r^2} dr + c rightarrow v^2 = frac{2gR^2}{r} + C) to find c we use at r=R
v=v_e thus we get (v_e^2 – frac{2gR^2}{R} = C)… where 2c=C=constant substituting value of c we get (v^2=frac{2gR^2}{r} + v_e^2 – 2gR)…..if r>>R (frac{2gR^2}{r} = 0) and particle to get escape from earth v≥0 –> v_e² – 2gR≥0 –> (v_e=sqrt{2gR}) to find v_e
from the moon g becomes (frac{g}{6}), R=2000km=2×10⁶ m, g=10ms^-2
therefore (v_e = sqrt{2*frac{10}{6}(2×10^6)} = 2.58kms^{-1}.).

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250+ TOP MCQs on Table of General Properties of Laplace Transform and Answers

Ordinary Differential Equations online quiz focuses on “Table of General Properties of Laplace Transform”.

1. Find the L(sin³ t).
a) (frac{3}{4(s^2+1)}-frac{1}{4(s^2+9)})
b) (frac{3}{4(s^2+1)}-frac{3}{4(s^2+9)})
c) (frac{3}{4(s^2+1)}-frac{9}{4(s^2+9)})
d) (frac{3}{4(s^2-1)}-frac{3}{4(s^2+9)})
Answer: b
Explanation: In the given question
= L(sin³ t)
= (L left (frac{3 sint}{4}right )-L left (frac{1}{4} sin⁡(3t)right ))
= (frac{3}{4(s^2+1)}-frac{3}{4(s^2+9)}).

2. Find the (L(e^{2t} (1+t)^2)).
a) (frac{1}{s-2}+frac{2}{(s-2)^3} + frac{2}{(s-2)^2})
b) (frac{⁡3}{s-2}+frac{2}{(s-2)^3} + frac{2}{(s-2)^2})
c) (frac{1}{s-2}+frac{2}{(s+2)^3} + frac{2}{(s-2)^2})
d) (frac{1}{s-2}+frac{2}{(s-2)^3} )
Answer: a
Explanation: In the given question,
(L((1+t)^2 )=L(1+t^2+2t))
(=frac{1}{s}+frac{2}{s^3}+frac{2}{s^2} )
(L(e^{2t} (1+t)^2)=frac{1}{s-2}+frac{2}{s-2^3} + frac{2}{(s-2)^2}).——————–By the first shifting property

3. Find the Laplace Transform of g(t) which has value (t-1)³ for t>1 and 0 for t<1.
a) (e^{-2as}×frac{6}{s^4})
b) (e^{-as}×frac{24}{s^5})
c) (e^{-as}×frac{6}{s^4})
d) (e^{-as}×frac{24}{s^4})
Answer: c
Explanation: In the given question,
We use the second shifting property.
Let f(t)=t³
(L(f(t))=frac{6}{s^4})
By the second shifting,
(L(g(t))=e^{-as}×frac{6}{s^4})
.

4. Find the L(t e^-2t sinh⁡(4t)).
a) (frac{8s+16}{(s^2+2s-12)^2})
b) (frac{2s+16}{(s^2+2s-12)^2})
c) (frac{8s+16}{(s^2+21s-12)^2})
d) (frac{8s+16}{(s^2+s-12)^2})
Answer: a
Explanation: In the given question,
L(t e^-2t sinh⁡(4t))
L(sinh⁡(4t))=(frac{4}{s^2-16})
By effect of multiplication of t
L(t×sinh⁡(4t))=((-1) frac{d}{ds} frac{4}{s^2-16})
L(t×sinh⁡(4t))=(frac{8s}{(s^2-16)^2})
By First shifting property
L(t e^-2t sinh⁡(4t))=(frac{8(s+2)}{((s+2)^2-16)^2} = frac{8s+16}{(s^2+2s-12)^2}).

5. Find the L(t+sin(2t)).
a) (frac{1}{s}+frac{2}{(s^2+4)})
b) (frac{1}{s}+frac{3}{(s^2+4)})
c) (frac{1}{s}+frac{2}{(s^2+2)})
d) (frac{2}{s}+frac{2}{(s^2+4)})
Answer: a
Explanation: In the given question,
L(t+sin⁡(2t))
=(frac{1}{s}+frac{2}{(s^2+4)}).

6. The L(te^-3t cos⁡(2t)cos⁡(3t)) is given by (kleft [frac{25-(s+3)^2}{((s+3)^2+25)^2} + frac{(1-(s+3)^2)}{((s+3)^2+1)^2}right ]). Find the value of k.
a) 0
b) 1
c) (frac{1}{2})
d) (frac{-1}{2})
Answer: d
Explanation: In the given question,
L(e^-3t cos⁡(2t)cos⁡(3t))
L(cos⁡(2t) cos⁡(3t))
=(frac{1}{2} L(cos⁡(5t)+cos⁡(t)))
=(frac{1}{2} left (frac{s}{(s^2+25)}right )+frac{1}{2} left (frac{s}{(s^2+1)}right ))
By effect of multiplication by t
L(t×cos⁡(2t) cos⁡(3t))=((-1)×frac{1}{2}×frac{d}{ds} frac{1}{2} left (frac{s}{(s^2+25)}right )+frac{1}{2} left (frac{s}{(s^2+1)}right ))
=(frac{-1}{2} left [frac{25-s^2}{(s^2+25)^2} + frac{(1-s^2)}{(s^2+1)^2}right])
By the effect of first shifting,
L(te^-3t cos⁡(2t)cos⁡(3t))=(frac{-1}{2} left [frac{25-(s+3)^2}{((s+3)^2+25)^2} + frac{1-(s+3)^2}{((s+3)^2+1)^2}right ])
k=(frac{-1}{2}).

7. Find the (Lleft (frac{sinh⁡(at)}{t}right )).
a) (frac{1}{2} log⁡ left (frac{s×a}{s-a}right ))
b) (frac{1}{2} log⁡ left (frac{s-a}{s+a}right ))
c) (frac{1}{2} log⁡ left (frac{s+a}{s-a}right ))
d) (frac{1}{3} log⁡ left (frac{s+a}{s-a}right ))
Answer: c
Explanation: In the given question,
L(sinh⁡(at))=(frac{a}{s^2-a^2})
By effect of division by t,
(Lleft (frac{sinh⁡(at)}{t}right )=int_{s}^{infty}frac{a}{s^2-a^2} ds)
=(a×frac{1}{2a}×log⁡ left (frac{s-a}{s+a}right ) ) in limits s to ∞
=(frac{1}{2} log⁡(0)-frac{1}{2} log⁡ left (frac{s-a}{s+a}right ))
=(frac{1}{2} log⁡ left (frac{s+a}{s-a}right )).

8. Find the (Lleft (frac{d}{dt}(frac{sin⁡t}{t})right)).
a) s×cot^-1 s-1
b) s×tan^-1 s-1
c) s×cot⁡(s)-1
d) s×tan⁡(s)-1
Answer: a
Explanation: In the given question,
L(sint)=(frac{1}{s^2+1})
By effect of division by t,
(L(frac{sin⁡t}{t})=int_{s}^{infty}frac{1}{s^2+1} ds)
=(frac{pi}{2}-tan^{-1}s)
=cot^-1s
By effect of derivative of Laplace Transform,
(Lleft (frac{d}{dt}left (frac{sin⁡t}{t}right )right )=s×cot^{-1}s-1).

9. Find the (L(int_{0}^{t}sin⁡(u) cos⁡(2u)du)).
a) (frac{1}{2s} left [frac{3}{s^2+9}-frac{1}{s^2+1}right ])
b) (frac{1}{2s} left [frac{9}{s^2+9}-frac{1}{s^2+1}right ])
c) (frac{1}{2s} left [frac{3}{s^2+9}+frac{1}{s^2+1}right ])
d) (frac{1}{s} left [frac{3}{s^2+9}-frac{1}{s^2+1}right ])
Answer: a
Explanation: In the given question,
L(sin⁡(t) cos⁡(2t))=(frac{1}{2} L(sin⁡(3t)-sin⁡(t)))
=(frac{1}{2} left [frac{3}{s^2+9}-frac{1}{s^2+1}right ])
By Laplace Transformation of an integral,
(L(int_{0}^{t}sin⁡(u) cos⁡(2u)du)=frac{1}{2s} left [frac{3}{s^2+9}-frac{1}{s^2+1}right ])

10. Which of the following is not a term present in the Laplace Transform of e^2t sin⁴ t.
a) (frac{3}{8s})
b) (frac{3}{8(s-2)})
c) (frac{s}{8((s-2)^2+16)})
d) (frac{s}{2((s-2)^2+4)})
Answer: a
Explanation: In the given question,
sin⁴ t=(frac{1+cos^2(2t)-2cos⁡(2t)}{4})
L(sin⁴ t)=(Lleft (frac{1+cos^2 (2t)-2 cos⁡(2t)}{4}right ))
=(frac{3}{8s}-frac{s}{2(s^2+4)}+frac{s}{8(s^2+16)})
By First Shifting Property
L(e^2t sin⁴ t)=(frac{3}{8(s-2)}-frac{s}{2((s-2)^2+4)}+frac{s}{8((s-2)^2+16)})

11. If ((erf⁡(sqrt{t}))=frac{1}{ssqrt{s}}), then what is (L(erf⁡(2sqrt{t})))?
a) (frac{2}{sqrt{s}})
b) (frac{1}{ssqrt{s}})
c) (frac{2}{ssqrt{s}})
d) (frac{4}{ssqrt{s}})
Answer: c
Explanation: In the given question,
By using scaling property of Laplace Transform,
(L(erf⁡(2sqrt{t}))=frac{1}{4}×frac{1}{frac{s}{4}×sqrt{frac{s}{4}}})
=(frac{2}{ssqrt{s}}).

12. Find the value of L(3^2t).
a) (frac{1}{s-2 log⁡(3)})
b) (frac{1}{s+2 log⁡(3)})
c) (frac{1}{s-3 log⁡(2)})
d) (frac{1}{s+3 log⁡(2)})
Answer: a
Explanation: In the given question,
The formula is,
L(c^at)=(frac{1}{s-a log(c)})
L(3^2t)=(frac{1}{s-2 log⁡(3)}).

Global Education & Learning Series – Ordinary Differential Equations.

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250+ TOP MCQs on Directional Derivative and Answers

Linear Algebra Objective Questions & Answers focuses on “Directional Derivative”.

1. Find the directional derivative of φ = xy² + yz³ at (1, -1, 1), towards the point (2, 1, -1).
a) ( frac{5}{3} )
b) ( frac{-5}{3} )
c) ( frac{7}{3} )
d) ( frac{1}{3} )
Answer: a
Explanation: (frac{dφ}{dx} = y^2, frac{dφ}{dy} =2xy+z^3, frac{dφ}{dz} = 3yz^3 )
(∇ φ = y^2 hat{i} + (2xy+z^3) hat{j} + 3yz^3hat{k} ̂)
([∇ φ]_{(1,-1,1)} = hat{i} – hat{j} – 3hat{k} ̂)
Now (a ̅ ) is along the line joining(1,-1,1) and (2,1,-1)
(a ̅ =(2-1) hat{i} + (1+1) hat{j} + (-1-1) hat{k} = hat{i} + 2hat{j} – 2hat{k} )
(a ̂ = frac{(i ̂ +2 j ̂ -2hat{k})}{sqrt{(1+4+4)}} = frac{1}{3} (i ̂ +2 j ̂ -2hat{k}))
∴ Directional derivative = (∇ φ. a ̂ = (i ̂ – j ̂ -3hat{k}) . frac{1}{3} (i ̂ +2 j ̂ -2hat{k}))
(= frac{1}{3} (1-2+6) = frac{5}{3} .)

2. The directional derivative of φ(x,y) at the point A(3,2) towards the point B(2,3). What is (3sqrt{2}) and toward the point (1,0) is (sqrt{8}). What is the directional derivative at the point A towards the point D.
a) (frac{6}{5})
b) (frac{7}{sqrt{5}} )
c) (frac{6}{sqrt{5}} )
d) (frac{7}{5} )
Answer: b
Explanation: Here (overrightarrow{AB} = (2-3) i ̂ + (3-2) j ̂ = – i ̂ + j ̂ )
Directional derivative of φ(x,y) toward (overrightarrow{AB} ) is
(∇ φ . widehat{(AB)} = (hat{i} frac{∂φ}{∂x}+ hat{j}frac{∂φ}{∂y}) .(frac{- i ̂ + j ̂ }{sqrt{2}}) = 3sqrt{2})
(– frac{∂φ}{∂x} + frac{∂φ}{∂y} = 6 …..(i) )
Directional derivative at A(3,2) towards C(1,0) is
(∇ φ . widehat{(AC)} = (i ̂ frac{∂φ}{∂x}+ j ̂ frac{∂φ}{∂y}) . frac{-2 i ̂-2j ̂ }{sqrt{8}} = sqrt{8} )
(– 2 frac{∂φ}{∂x} -2 frac{∂φ}{∂y} = 8 ,or, frac{∂φ}{∂x} + frac{∂φ}{∂y} = 4 …..(ii) )
From (i) & (ii), (∇ φ = -5hat{i} + hat{j} )
Hence directional derivative at A(3,2) towards D(2,4) is
(∇ φ . widehat{AD} = (-5hat{i} + hat{j}) . left(frac{-5hat{i} + hat{j}}{sqrt{5}}right) = frac{7}{sqrt{5}}.)

3. For the function f = x²y + 2y²x, at the point P(1,3), what is the direction in which the directional derivative is zero?
a) (-13hat{i} – 24hat{j} )
b) (13hat{i} + 24 hat{j} )
c) (±13hat{i} ∓24hat{j} )
d) (∓13hat{i} ± 24hat{j} )
Answer: c
Explanation: Let the directional derivative is zero along (a1hat{i} + a2hat{j})
(Then ∇ f = hat{i} frac{∂f}{∂x} +hat{j} frac{∂f}{∂y} = hat{i} (2xy+2y^2) + hat{j}(x^2+4xy) )
(i.e.[∇ f]_{(1,3)} = 24hat{i} + 13hat{j} )
([∇ f]_{(1,3)} . (a_1hat{i} +a_2hat{j}) = 0 )
((a_1hat{i} +a_2hat{j}) . (a_1hat{i} + a_2hat{j}) = 0 )
(24a_1 + 13a_2 = 0 ∴ frac{a1}{13}=frac{(-a2)}{24} )
∴ Direction are (overline{u1} = 13hat{i} – 24hat{j}, overline{u2} = -13hat{i} + 24hat{j}. )

4. The unit normal vector n ̂ of the cone of revolution z² = 4(x² + y²) at the Point P (1, 0, 2) is?
a) ([±frac{2}{sqrt{5}}, 0, ∓frac{1}{sqrt{5}}] )
b) ([frac{2}{sqrt{5}}, 0, frac{1}{sqrt{5}}] )
c) ([-frac{2}{sqrt{5}}, 0, -frac{1}{sqrt{5}}] )
d) ([∓frac{2}{sqrt{5}}, 0, ±frac{1}{sqrt{5}}] )
Answer: a
Explanation: A cone is a level surface say f = 0 of f (x, y, z) = 4(x²+y²) – z²
(∴ ∇ f = 8xhat{i} + 8yhat{j} – 2zhat{k} )
or ( ∇ f_{(1,0,2)} = 8hat{i} – 4hat{k} )
(hat{n} = frac{(8hat{i} – 4hat{k})}{sqrt{(8)2+(-4)2}} )
( = frac{(8hat{i} – 4hat{k})}{sqrt{80}} )
( = frac{(8hat{i} – 4hat{k})}{sqrt{(16*5)}} = frac{(2hat{i} – hat{k})}{sqrt{5}} , or , hat{n} = [frac{2}{sqrt{5}}, 0, frac{-1}{sqrt{5}}] )
There are two possible unit vectors, if one is along direction then (hat{n} ) other will be along – (hat{n} )
(∴ hat{n} = [±frac{2}{sqrt{5}}, 0, ∓frac{1}{sqrt{5}}]. )

5. For what value of a & b for which the two surfaces ax² – byz = (a+2)x & 4x²y + z³ = 4 will be orthogonal to each other at the point (1,-1,2).
a) (a = 1 , & , b = frac{3}{2} )
b) (a = frac{-3}{2} , & , b = 1 )
c) (a = frac{3}{2} , & , b = 1 )
d) (a = frac{-3}{2} , & , b = -1 )
Answer: c
Explanation: Let (f_1 = ax^2 – byz – (a+2)x , & , f_2= 4x^2y+z^3 – 4 )
(∴ ∇ f_1 = [2ax – (a+2)]hat{i} – (-bz)hat{j} + (-bz)hat{k} )
also ( ∇ f_1|_{(1,-1,2)} = (a-2) hat{i} + bhat{j} + bhat{k} )
And ( ∇ f_2 = (8xy) hat{i} + (4x^2)hat{j} + (3z^2)hat{k}, also , ∇ f_2|_{(1,-1,2)} = -8hat{i} + 4hat{j} + 12hat{k} )
Now, if the two surfaces cut each other orthogonally at point P(1,-1,2)
then,
(∇ f_1 . ∇ f_2 = 0 )
Or ( [(a-2)hat{i} + bhat{j} + bhat{k}] . (-8hat{i} + 4hat{j} + 12hat{k}) = 0
= -2a + b = -4 …..(i) )
Also if point P lies on both the curve hence it would also satisfy
Both the curves, i.e substituting P(1,-1,2) in equation
(= ax^2-byz = (a+2) )
(= a+2b = a+2 ⇨ b=1 )
From eqn (i) (-2a+1 = -4 => 2a = 3 => a = frac{3}{2}. )

6. The temperature of a point in space is given by T = x² + y² – z. An insect located at a point (1, 1, 2) desire to fly in such a direction such that it will get warm as soon as possible. In what direction it should move?
a) (frac{-2hat{i}}{3}+frac{2hat{j}}{3}+frac{-hat{k}}{3} )
b) (frac{2hat{i}}{3}+frac{-2hat{j}}{3}+frac{-hat{k}}{3} )
c) (frac{2hat{i}}{3}+frac{2hat{j}}{3}+frac{hat{k}}{3} )
d) (frac{2hat{i}}{3}+frac{2hat{j}}{3}+frac{-hat{k}}{3} )
Answer: d
Explanation: Here the insect will get warm quickly if it moves normal to the given
Surface in space.
As (T = x^2+y^2-z )
(∴ ∇ T = (frac{∂}{∂x}hat{i} + frac{∂}{∂x}hat{j} + frac{∂}{∂x}hat{k})(x^2+y^2-z) )
( = 2xhat{i} + 2yhat{j} – hat{k} )
Now ( ∇ T |_{(1,1,2)} = 2hat{i} + 2hat{j} – hat{k} )
∴ required direction( = frac{∇ T}{|∇ T|} )
(=frac{2hat{i} + 2hat{j} – hat{k}}{sqrt{(4+4+1)}} )
(= frac{2hat{i} +2hat{j} – hat{k}}{3} )
(= frac{2hat{i}}{3}+frac{2hat{j}}{3}+frac{-hat{k}}{3}. )

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250+ TOP MCQs on Solution of 1D Heat Equation and Answers

Fourier Analysis Interview Questions and Answers for Experienced people focuses on “Solution of 1D Heat Equation”.

1. The partial differential equation of 1-Dimensional heat equation is ___________
a) u_t = c²u_xx
b) u_t = pu_xx
c) u_tt = c²u_xx
d) u_t = – c²u_xx
View Answer

Answer: a
Explanation: The one-dimensional heat equation is given by u_t = c²u_xx where c is the constant and u_t represents the one time partial differentiation of u and u_xx represents the double time partial differentiation of u.

2. When using the variable separable method to solve a partial differential equation, then the function can be written as the product of functions depending only on one variable. For example, U(x,t) = X(x)T(t).
a) True
b) False
View Answer

Answer: a
Explanation: When solving a partial differential equation using a variable separable method, then the function can be written as the product of functions depending on one variable only.

3. The one dimensional heat equation can be solved using a variable separable method. The constant which appears in the solution should be __________
a) Positive
b) Negative
c) Zero
d) Can be anything
View Answer

Answer: b
Explanation: Since the problems are dealing on heat conduction, the solution must be a transient solution. Therefore the constant should be negative, i.e., k = – p².

4. When solving the 1-Dimensional heat equation for the conduction of heat along the rod without radiation with conditions:
i) u(x,t) is finite for t tends to infinite
ii) u_x(0,t) = 0 and u_x(l,t) = 0
iii) u(x,t) = x(l-x) for t=0 between x=0 and x=l, which condition is the best to use in the first place?
a) u_x(0,t) = u_x(l,t) = 0
b) u(x,t) = x(l-x) for t=0 between x=0 and x=l.
c) u(x,t) = x(l-x) for x=0 between t=0 and t=l.
d) u(0,t) = u(l,t) = 0
View Answer

Answer: a
Explanation: Boundary conditions are always used first to solve the partial differential equations. Using these boundary conditions, we can remove one constant thus making only one constant remaining to remove. The last constant is removed using the initial conditions.

5. Solve the 1-Dimensional heat equation for the conduction of heat along the rod without radiation with conditions:
i) u(x,t) is finite for t tends to infinite
ii) u_x(0,t) = 0 and u_x(l,t) = 0
iii) u(x,t) = x(l-x) for t=0 between x=0 and x=l.
a) U(x,t) =(frac{l^2}{3}/2 + ∑cos⁡(frac{nπx}{l}) e^{frac{-c^2 n^2 π^2 t}{l^2}} frac{-4l^2}{(2m)^2+π^2} )
b) U(x,t) =(frac{l^2}{3} + ∑cos⁡(frac{nπx}{l}) e^{frac{-c^2 n^2 π^2 t}{l^2}} frac{-4l^2}{(2m)^2+π^2} )
c) U(x,t) =(frac{l^2}{3} + ∑cos⁡(frac{nπx}{l}) e^{frac{-c^2 n^2 π^2 t}{l^2}} frac{4l^2}{(2m)^2+π^2} )
d) U(x,t) =(frac{l^2}{3}/2 + ∑cos⁡(frac{nπx}{l}) e^{frac{-c^2 n^2 π^2 t}{l^2}} frac{4l^2}{(2m)^2+π^2} )
View Answer

Answer: a
Explanation: u(x,t) = (c cospx + c’ sinpx) (c’’ e^-p² c²t)
u_x = (-cp sinpx + c’p cospx) (c’’ e^-p² c²t)
Applying the first condition of second condition,
C’ = 0
Now applying the second condition of the second condition,
(p= frac{nπ}{l} )
Now we have only one constant left. This can be solved using the third condition.
U(x,t) =( frac{a_0}{2} + ∑cos⁡(frac{nπx}{l}) a_n e^{frac{-c^2 n^2 π^2 t}{l^2}} )
(a_0 = frac{2}{l} ∫_0^l x(l-x)dx = frac{l^2}{3} )
(a_n = frac{2}{l} ∫_0^l x(l-x) cosnx dx = frac{-4l^2}{(2m)^2+π^2}. )
(U(x,t) = frac{l^2}{3}/2 + ∑cos⁡(frac{nπx}{l}) e^{frac{-c^2 n^2 π^2 t}{l^2}} frac{-4l^2}{(2m)^2+π^2}.)

6. A rod of 30cm length has its ends P and Q kept 20°C and 80°C respectively until steady state condition prevail. The temperature at each point end is suddenly reduced to 0°C and kept so. Find the conditions for solving the equation.
a) u(0,t) = 0 = u(30,t) and u(x,0) = 20 + 60/10 x
b) u_x(0,t) = 0 = u_x(30,t) and u(x,0) = 20 + 60/30 x
c) u_t(0,t) = 0 = u_t(30,t) and u(x,0) = 20 + 60/10 x
d) u(0,t) = 0 = u(30,t) and u(x,0) = 20 + 60/30 x
View Answer

Answer: d
Explanation: 0 and 30 are the end points. So, at these points the function is zero. Hence u(0,t) = 0 = u(30,t). Next due to steady state conditions, at the beginning that is initial conditions, we have u(x,0) = 20 + 60/30 x.

7. Is it possible to have a solution for 1-Dimensional heat equation which does not converge as time approaches infinity?
a) Yes
b) No
View Answer

Answer: b
Explanation: It is not possible to have a solution which does not converge as time approaches infinity because the solution to a heat equation must be transient.

8. Solve the equation u_t = u_xx with the boundary conditions u(x,0) = 3 sin (nπx) and u(0,t)=0=u(1,t) where 0<x<1 and t>0.
a) (3∑_{n=1}^∞ ) e^{-n² π² t} cos⁡(nπx)
b) (∑_{n=1}^∞ ) e^{-n² π² t} sin⁡(nπx)
c) (3∑_{n=1}^∞ ) e^{-n² π² t} sin⁡(nπx)
d) (∑_{n=1}^∞ ) e^{-n² π² t} cos(nπx)
View Answer

Answer: c
Explanation: u(x,t) = (c cospx + c’ sinpx) (c’’ e^-p²t)
When x=0, c=0 and when x=1, p=nπ.
When t=0, 3 sin (nπx) = (∑_{n=1}^∞ ) b_n e^{-n² π² t} sin⁡(nπx)
Therefore b_n =3 for all n
Hence the solution is (3∑_{n=1}^∞ ) e^{-n² π² t} sin⁡(nπx).

9. If two ends of a bar of length l is insulated then what are the conditions to solve the heat flow equation?
a) u_x(0,t) = 0 = u_x(l,t)
b) u_t(0,t) = 0 = u_t(l,t)
c) u(0,t) = 0 = u(l,t)
d) u_xx(0,t) = 0 = u_xx(l,t)
View Answer

Answer: a
Explanation: Since the ends are insulated no heat can flow through the ends of the bar. Therefore u_x(0,t) = 0 = u_x(l,t).

10. The ends A and B of a rod of 20cm length are kept at 30°C and 80°C until steady state prevails. What is the condition u(x,0)?
a) 20 + ⁵⁄₂ x
b) 30 + ⁵⁄₂ x
c) 30 + 2x
d) 20 + 2x
View Answer

Answer: b
Explanation: u_xx=0.
The solution to this equation is u=a+bx. Since at x=0, u=30 and at x=20, u=80,
a = 30 and b = (frac{(80-30)}{20} = frac{5}{2} )
Therefore, u= 30 + (frac{5}{2} x).

Global Education & Learning Series – Fourier Analysis.

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