- #1

- 327

- 1

## Homework Statement

Tan∅ dr + 2r d∅=0

## Homework Equations

## The Attempt at a Solution

∫1/2r dr +∫1/tan∅ d∅=0

1/2ln(2r)+ln(tan∅)=c

ln[r(tan∅)]=ln(c)

r(tan∅)=c

the solution in the book says the answer is rsin^2∅=c

Where did I go wrong?

Thank you

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- Thread starter Mdhiggenz
- Start date

- #1

- 327

- 1

Tan∅ dr + 2r d∅=0

∫1/2r dr +∫1/tan∅ d∅=0

1/2ln(2r)+ln(tan∅)=c

ln[r(tan∅)]=ln(c)

r(tan∅)=c

the solution in the book says the answer is rsin^2∅=c

Where did I go wrong?

Thank you

- #2

- 3,816

- 92

Here's your mistake:

∫1/tan∅ d∅≠ln(tan∅)

∫1/tan∅ d∅≠ln(tan∅)

- #3

- 327

- 1

Wow I see it would be ∫cos∅/sin∅ d∅ and with some u substitution it would give us ln(sin∅) correct?Here's your mistake:

∫1/tan∅ d∅≠ln(tan∅)

- #4

- 3,816

- 92

Wow I see it would be ∫cos∅/sin∅ d∅ and with some u substitution it would give us ln(sin∅) correct?

Yep, that's right but you have done one more mistake.

∫1/2r dr≠(1/2)ln(2r). It is equal to (1/2)ln(r).

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