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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
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Assignment No. 1
MTH 301 (Spring 2010)
Maximum Marks: 25
Due Date: 28^{th} 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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