Question Bank | unit 2 multivariable calculus

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Question Bank

UNIT–2

Multivariable Calculus

Question-1: Test the continuity of the following function-

Sol. (1) the function is well defined at (0,0)

(2) check for the second step,

That means the limit exists at (0,0)

Now check step-3:

So that the function is continuous at origin.

Question-2: Evaluate

Sol . 1.

Here f1 = f2

3. Now put y = mx, we get

Here f1 = f2 = f3

Now put y = mx²

Therefore ,

F1 = f2 = f3 =f4

We can say that the limit exists with 0.

Question-3: Obtain all the second order partial derivative of the function:

f( x, y) = ( x³y² - xy⁵)

Sol. 3x²y² - y⁵, 2x³y – 5xy⁴,

= = 6xy²

= 2x³ - 20xy³

= = 6x²y – 5y⁴

= = 6x²y - 5y⁴

Question-4: if , then show that-

Sol. Here we have,

u = …………………..(1)

Now partially differentiate eq.(1) w.r to x and y , we get

………………..(2)

And now,

………………….(3)

Adding eq. (1) and (3) , we get

= 0

Hence proved.

Question-5: if w = x² + y – z + sint and x + y = t, find

(a) y,z

(b) t, z

Sol. With x, y, z independent, we have

t = x + y, w = x² + y - z + sin (x + y).

Therefore,

y,z = 2x + cos(x+y)(x+y)

= 2x + cos (x + y)

With x, t, z independent, we have

Y = t-x, w= x² + (t-x) + sin t

thus t, z = 2x - 1

Question-6: If z is the function of x and y , and x = , y = , then prove that,

Sol. Here , it is given that, z is the function of x and y & x , y are the functions of u and v.

So that,

……………….(1)

And,

………………..(2)

Also there is,

x = and y = ,

Now,

, , ,

From equation(1) , we get

……………….(3)

And from eq. (2) , we get

…………..(4)

Subtracting eq. (4) from (3), we get

= ) – (

= x

Hence proved.

Question-7: Find the equation of the tangent plane and normal line to the surface

Sol.

Here,

At point (1, 2, -1)-

Therefore the equation of the tangent plane at (1, 2, -1) is-

Equation of normal line is-

Question-8: Find the point on plane nearest to the point (1, 1, 1) using Lagrange’s method of multipliers.

Solution:

Let be the point on sphere which is nearest to the point . Then shortest distance.

Let

Under the condition … (1)

By method of Lagrange’s undetermined multipliers we have

… (2)

… (3)

i.e. &

… (4)

From (2) we get

From (3) we get

From (4) we get

Equation (1) becomes

i.e.

y = 2

Question-9: Evaluate , where dA is the small area in xy-plane.

Sol. Let , I =

= 84 sq. Unit.

Which is the required area.

Question-10: Evaluate

Sol. Let us suppose the integral is I,

I =

Put c = 1 – x in I, we get

I =

Suppose , y = ct

then dy = c

Now we get,

I =

As we know that by beta function,

Which gives,

Now put the value of c, we get

Question-11: Evaluate ey/x dy dx.

Soln. :

Given : I = ey/x dy dx

Here limits of inner integral are functions of y therefore integrate w.r.t y,

I = dx

I =

= =

ey/x dy dx=

Question-12: Evaluate

Soln. :

Let,I =

Here limits for both x and y are constants, the integral can be evaluated first w.r.t any of the variables x or y.

I = dy

= 

Question-13: Evaluate  over x  1, y 

Soln. :

1a1 Let I= over x  1, y 

The region bounded by x  1 and y 

Is as shown in Fig. 6.3.

Fig. 6.3

Take a vertical strip along strip x constant and y varies from y =

To y = . Now slide strip throughout region keeping parallel to y-axis. Therefore y constant and x varies from x = 1 to x = .

= [ ∵ dx = tan–1 (x/a)]

= =

=– = (0 – 1)

 I =

Question-14: Evaluate

Solution: Let

I =

(Assuming m = )

= dxdy

= dx

I =

Question-15: Find Volume of the tetrahedron bounded by the co-ordinates planes and the plane

Solution: Volume = ………. (1)

Put ,

From equation (1) we have

V =

=24

=24 (u+v+w=1) By Dirichlet’s theorem.

=24

= = = 4

Volume =4

Question-16: Evaluate

Soln. :

The region of integration bounded by

y = 0, y = and x = 0, x = 1

y =  x2 + y2 = x

The region bounded by these is as shown in Fig. 6.8.

Convert the integration in polar co-ordinates by using x = r cos , y = r sin  and dx dy = r dr d

 x2 + y2= x becomes r = cos 

y = 0 becomes r sin  = 0  = 0

x = 0 becomes r cos  = 0  =

And x = 1 becomes r = sec 

Take a radial strip SR with angular thickness , Along strip  constant and r varies from r = 0 to r = cos . Turning strip throughout region therefore  varies from = 0 to  =

 I = r dr d

= 4 cos  sin  d r dr

= 4 cos  sin  d [–]

= – 2 cos  sin  [+1] d

= – 2 [cos  sin  – cos  sin ] d

= –2 + 2 cos  sin  d

= – + 2

= + 1 =

Question-17: Sketch the area of double integration and evaluate

dxdy

Soln. :

Let I =dxdy

The region of integration is bounded by the curves

x = y, x = and y = 0, y = Fig. 6.9

i.e. x = y, x2 + y2 = a2 and y = 0, y =

The region bounded by these is as shown in Fig. 6.9.

The point of intersection of x = y and x2 + y2 = a2 is x =

Convert given integration in polar co-ordinates by using polar transformation x = r cos , y = r sin  and dx dy = r dr d

 x = y gives r cos  = r sin  tan  = 1  =

x2 + y2 = a2 r2 = a2 r =  a

y = 0 gives r sin  = 0  = 0.

y = gives r sin  =  r =  cosec 

Take a radial strip SR, along SR  constant and r varies from r = 0 to r = a. Turning this strip throughout region therefore  varies from  = 0 to  =

 I = log r2 r dr d = 2 d r log r dr

I =

= 2 d

= 2 []

I = [/4] =

Question-18: Find the area between the curves and its asymptote.

Solution: The curve is symmetrical about x axis not passing through origin. Also is the asymptote to the curve and intersect x axis at for curve doesn’t exists. And for and. Because of symmetry