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Assignment No. 1 MTH 301 (Spring 2010)...Not the Solution.....But Formulas And Examples for Solving Assignment. Just Put the Values of your Assignment in Attach File Formulas and Solve.

Download the attached file pls

 

Spring 2010_MTH301_1.doc

 

 

________________________________________

Assignment No. 1

MTH 301 (Spring 2010)

                                           

Maximum Marks: 25

                                                Due Date: 28th April 2010  

 

 

 

 

 

 

Spherical co-ordinates of a point are?

Spherical co-ordinates of a point are .(4,3.14/2,3.14/6)
a. Convert spherical co-ordinates to Rectangular co-ordinates
b. Convert spherical co-ordinates to cylindrical co-ordinates.
c. Verify your answer by converting back spherical co-ordinates from any one of these, that is, either from rectangular co-ordinates or cylindrical co-ordinates.

 

Answer

 

Assuming the spherical coordinates are (r,theta,phi), rectangular coordinates are:
x=r sin(theta)cos(phi)=4 * sin(3.14/2) cos(3.14/6)=3.464
y=r sin(theta)sin(phi)=4 * sin(3.14/2) sin(3.14/6)=2
z=r cos(theta)=4 * cos(3.14/2)=0

Cylindrical coordinates are:
rho = r sin(theta)=4 * sin(3.14/2)=4
phi = phi=3.14/6
z= r cos (theta)=4 * cos(3.14/2)=0

To convert back from cylindrical coordinates
r=sqrt{rho^2+z^2}=rho=4
theta=atan2 {rho,z}=atan2(4,0)=3.14/2
phi=phi=3.14/6

 

 

Question Details:

Rectangular co-ordinates of a point are

a.     
Convert Rectangular co-ordinates to Spherical
co-ordinates

 

b.     Convert
Rectangular co-ordinates to Cylindrical co-ordinates.

 

c.     
Verify your answer by converting back Rectangular
co-ordinates from any one of these, that is, either from Spherical co-ordinates
or Cylindrical co-ordinates.

Response Details:

 

a)

So to go from Rec to Spherical you need to know that

Cartesian is in the form(x,y,z) and Spherical are in (r,j,q)

so to convert your x  your r is equal to  so (1^2)+(1^2)+(-2√2) which equals 10 then you sqrt that

and r =

then to find q use formula tanθ=y/x and solve for θ so  which equals 45 degrees

q=45 degrees

and to find j know that z = r*cos j so solve for j

(-2√2) = (√10 )* cos j

(-2√2)/(√10) = j =153.4349 degrees

so your complete new point is

b) To convert from rec to Cylindrical

Cartesian is (x,y,z) and Cyl. is (r,q,z) notice that they have z in common so your z remains the same no change

use x to find r with the equation

so (1^2)+(1^2) = 2 then sqrt that now       r=√2

find q using the same method as above which is the inversetan of y/x

θ=invtan(1/1)= 45    q=45

and z is the same. your new point is

c) I believe it says to only convert 1 backwards so to go from Cyn to Rec

z stays the same

x=rcosq

y=rsinq

x=√2 cos45=1

y=√2 sin45=1

so (1,1,-2√2)

 

 

 

Question Details:

Describe the set of all points in xyz-coordinate
system at which f is continuous.

 

 

 

 

 

Answer :

 

 is continuous on the set of . is continuous on the set of . The product of two continuous functions is continuous on the intersections of their domains. Therefore,  is continuous on the set , which can further be described as

 

 

 

 

 

 

 

By considering different path approach, find whether?

By considering different path approach, find whether

Lim
(x,y)
(0,0)

xy
---------------
3x^2 + 2y^2

Exist or not ?

 

Answer

 

 

Substituting the limits to the function will give us 0 / 0, which is indeterminate: That is:

(0)(0) / [3(0)^2 + 2(0)^2] = 0 / 0

We need to use the Lhospital's rule considering two cases: For the limit of x and for the limit of y.. Thus, when we differentiate both the numerator and the denominator, we differentiate it using partial differentiation...

With respect to x (y is constant), the partial derivative of the function becomes:

y / [6x].

Substituting the limit, we get, 0 / 0. We apply LHospital's Rule (partial differentiation) again with respect to x (since we do the differentiation with respect to x from the start):


0 / 6 = 0. Take note of this 0.



Let's consider applying Lhospitals rule with respect to y. We have:


x / [4y]. Still 0 / 0, indeterminate. Applying the rule again wrt y:


= 0.

Thus, the limit exists.



To check on this answer, since our limit values are 0, lets remove the numerical coefficients:


xy / [x^2 + y^2].


Let x = y = 0.000000000001.

You will see that the denominator becomes very much larger than the numerator... A number divided by a very large number (almost infinity) is approximately equal to 0.

 

 

 

 

 

 

 


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yar koi meri hlp karay.. bhai ye assignment solv kar do mujay..ais main jo values ghalt hain wo solv kar do thek say..plz..aj last day hay

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