Differential and Integral Calculus Multiple Choice Questions on “Jacobians”.
1. The jacobian of p,q,r w.r.t x,y,z given p=x+y+z, q=y+z, r=z is ________
a) 0
b) 1
c) 2
d) -1
Answer: b
Explanation: We have to find
(J = frac{∂(p,q,r)}{∂(x,y,z)} = begin{vmatrix}
frac{∂p}{∂x} & frac{∂p}{∂y} &frac{∂p}{∂z}\
frac{∂q}{∂x} &frac{∂q}{∂y} &frac{∂q}{∂z}\
frac{∂r}{∂x} &frac{∂r}{∂y} &frac{∂r}{∂z}\
end{vmatrix})
But p=x+y+z, q=y+z, r=z (taking partial derivative)
(J=begin{vmatrix}
1&1&1\
0&1&1\
0&0&1\
end{vmatrix}(frac{∂p}{∂x}=1, frac{∂p}{∂y}=1, frac{∂p}{∂z}=1, frac{∂q}{∂x}=0, frac{∂q}{∂y}=1, frac{∂q}{∂z}=1, frac{∂r}{∂x}=0, \
frac{∂r}{∂y}=0, frac{∂r}{∂z}=1))
On expanding we get
J = 1(1 – 0) = 1
Thus j = 1.
2. Given (u=frac{yz}{x}, v=frac{zx}{y}, w=frac{xy}{z}) then the value of (frac{∂(u,v,w)}{∂(x,y,z)}) is ________
a) 4
b) -4
c) 0
d) 1
Answer: a
Explanation: By Data (u=frac{yz}{x}, v=frac{zx}{y}, w=frac{xy}{z})
(J = frac{∂(u,v,w)}{∂(x,y,z)} = begin{vmatrix}
frac{∂u}{∂x} & frac{∂u}{∂y} &frac{∂u}{∂z}\
frac{∂v}{∂x} &frac{∂v}{∂y} &frac{∂v}{∂z}\
frac{∂w}{∂x} &frac{∂w}{∂y} &frac{∂w}{∂z}\
end{vmatrix} = begin{vmatrix}
frac{-yz}{x^2} & frac{z}{x} &frac{y}{x}\
frac{z}{y} &frac{-zx}{y^2} &frac{x}{y}\
frac{y}{z} &frac{x}{z} &frac{-xy}{z^2}\
end{vmatrix})
(=frac{-yz}{x^2}Big{(frac{-zx}{y^2})(frac{-xy}{y^2}) – (frac{x}{z})(frac{x}{y})Big} – (frac{z}{x})Big{(frac{z}{y})(frac{-xy}{z^2})-(frac{y}{z})(frac{x}{y})Big} \
+ frac{y}{x}Big{(frac{z}{y})(frac{x}{z})-(frac{y}{z})(frac{-zx}{y^2})Big})
(=frac{-yz}{x^2} Big{frac{x^2}{yz} – frac{x^2}{yz}Big} – frac{z}{x}Big{frac{-x}{z} – frac{x}{z}Big} + frac{y}{x}Big{frac{x}{y} + frac{x}{y}Big} = 0 + 1 + 1 + 1 = 4)
Therefore (frac{∂(u,v,w)}{∂(x,y,z)} = 4.)
3. If u=x+3y2-z3, v=4x2 yz, w=2z2-xy then (frac{∂(u,v,w)}{∂(x,y,z)} ) at (1,1,1).
a) -184
b) -90
c) 20
d) 40
Answer: a
Explanation: Given that u=x+3y2-z3, v=4x2 yz, w=2z2-xy
(frac{∂(u,v,w)}{∂(x,y,z)} = begin{vmatrix}
frac{∂u}{∂x} & frac{∂u}{∂y} &frac{∂u}{∂z}\
frac{∂v}{∂x} &frac{∂v}{∂y} &frac{∂v}{∂z}\
frac{∂w}{∂x} &frac{∂w}{∂y} &frac{∂w}{∂z}\
end{vmatrix} = begin{vmatrix}
1 & 6y & -3z^2\
8xyz &4x^2 z &4x^2 y\
-y &-x &4z\
end{vmatrix})
(frac{∂(u,v,w)}{∂(x,y,z)} ,at, (1,1,1) = begin{vmatrix}
1 & 6 & -3\
8 &4&4\
-1 &-1 &4\
end{vmatrix})
=1(16+4) – 6(32+4) – 3(-8+4) = -184.
4. If x=rcosθ, y=rsinθ then the value of (frac{∂(x,y)}{∂(r,θ)}) is ________
a) 1
b) 0
c) r
d) (frac{1}{r})
Answer: c
Explanation: Wkt, (frac{∂(x,y)}{∂(r,θ)} = begin{vmatrix}
frac{∂x}{∂r} & frac{∂x}{∂θ}\
frac{∂y}{∂r} & frac{∂y}{∂θ}\
end{vmatrix} ,but, x=rcosθ, y=rsinθ)
Thus ( begin{vmatrix}
cosθ & -rsinθ\
sinθ & rcosθ\
end{vmatrix} = rcos^2 θ + rsin^2 θ = r.)
5. The application of Jacobians is significant in the evaluation of double integral of the form ∬f(x,y) dx dy and triple integral of the form ∭f(x,y,z)dx dy dz by transformation from one system of coordinate to the other.
a) True
b) False
Answer: a
Explanation: The principle of evaluation is analogous with the evaluation of ∫f(x)dx by taking a suitable substitution.
6. If u+v=ex cosy and u-v=ex sin y the value of (J(frac{u,v}{x,y})) is ________
a) e2x
b) (frac{e^2x}{2} )
c) (frac{-e^2x}{2} )
d) 0
Answer: c
Explanation: wkt (J(frac{u,v}{x,y}) = frac{∂(u,v)}{∂(x,y)} = begin{vmatrix}
frac{∂u}{∂x} & frac{∂u}{∂y}\
frac{∂v}{∂x} & frac{∂v}{∂y}\
end{vmatrix})
But we have to find u and v first using given equation
u+v=ex cos y………(1)
u-v=ex sin y……..(2)
solving (1)&(2)
we get u=(frac{e^x}{2} )(cosy+siny)
v=(frac{e^x}{2} )(cosy-siny)
(frac{∂u}{∂x} = frac{e^x}{2}(cosy+siny) )
(frac{∂v}{∂x} = frac{e^x}{2}(cosy-siny) )
(frac{∂u}{∂y} = frac{e^x}{2}(cosy-siny) )
(frac{∂v}{∂y} = frac{e^x}{2}(-cosy-siny) )
(J(frac{u,v}{x,y}) = begin{vmatrix}
frac{e^x}{2}(cosy+siny) & frac{e^x}{2}(cosy-siny)\
frac{e^x}{2}(cosy-siny) & frac{e^x}{2}(-cosy-siny)\
end{vmatrix})
((frac{e^x}{2})(frac{e^x}{2})){-(cosy+siny)2 -(cosy-siny)2}…….(expanding & solving by taking cos2 y+sin2 y=1)
=(frac{-e^2x}{2} ).
7. Which among the following is the definition of Jacobian of u and v w.r.t x and y?
a) (J(frac{x,y}{u,v}))
b) (J(frac{u,v}{x,y}))
c) (frac{∂(x,y)}{∂(u,v)} )
d) (frac{∂(u,x)}{∂(v,y)} )
Answer: b
Explanation: (J(frac{u,v}{x,y}) = frac{∂(u,v)}{∂(x,y)} = begin{vmatrix}
frac{∂u}{∂x} & frac{∂u}{∂y}\
frac{∂v}{∂x} & frac{∂v}{∂y}\
end{vmatrix})
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