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# MTH 301 Calculus II . Assignment No 2. Virtual University . Due Date Jul 17,2014.

Total marks: 20

Lecture # 23 to 30

Due date: July 17, 2014

DON’T MISS THESE Important instructions:

• All students are directed to use the font and style of text as is used in this document.
• This is an individual assignment, not group assignment, so keep in mind that you are supposed to submit your own, self made & different assignment even if you discuss the questions with your classmates. All similar assignments (even with some meaningless modifications) will be awarded zero marks and no excuse will be accepted. This is your responsibility to keep your assignment safe from others.
• Solve the assignment on MS word document and upload your word (.doc) files only

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### Replies to This Discussion

Arshad Bhai, In fourth line of Q3 correction is required.......

Equation of the curve in polar co-ordinates is r2=-4sin2θ

(i) Symmetry about the Initial Line.
r2=-4sin2θ
Remember that sine is an odd function
Put (r,-θ) in equation
r2=-4sin2 (-θ)
r2=-4sin (-2θ)
r2=4sin2θ
This equation is not symmetry about initial line
r2=-4sin2θ
Put (-r, θ) in equation
(-r) 2 =-4sin2θ
r2=-4sin2θ
This equation is symmetry about pole

rdrdθ ko dudθ main q convert kia haiiiiiiiiiiiiii???????????????????

orye question bilkul thk hai....???

Q:1:- Equation of the curve in polar co-ordinates is =-4sin2 discuss the symmetry of graph of this curve about.

1. Initial line
2. Pole

Equation of the curve in polar co-ordinates is

=-4sin2θ

1. Symmetry about the Initial Line.
=-4sin2θ
sin is odd function.
Put (r, -θ) in equation
=-4sin2 (-θ)
=-4sin (-2θ)
=4sin2θ
this equation is not symmetry about initial line
=-4sin2θ
Put (-r, θ) in equation
=-4sin2θ
=-4sin2θ
This equation is symmetry about pole

Check it

Attachments:

Q:1:- Equation of the curve in polar co-ordinates is =-4sin2 discuss the symmetry of graph of this curve about.
Initial line
Pole

Equation of the curve in polar co-ordinates is
=-4sin2θ
=-4sin2θ
sin is odd function.
Put (r, -θ) in equation
=-4sin2 (-θ)
=-4sin (-2θ)
=4sin2θ
this equation is not symmetry about initial line
=-4sin2θ
Put (-r, θ) in equation
=-4sin2θ
=-4sin2θ
This equation is symmetry about pole

MTH301_Complete_Solution

Attachments:

Tariq Bro, you have uploaded two solution for 3rd question which one is correct?

Doc file solution or above mentioned?

Plz ans.....

1

2

3

4

5

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