- #1

JoeyBob

- 256

- 29

- Homework Statement:
- See attached

- Relevant Equations:
- attached

So I start by isolating v

the speed here would be the square root of the partial t derivative divided by the sum of the partial x and y derivatives.

the amplitude, phi and the cos portion of the partial derivatives would all cancel out.

What I am left with is the sqrt(43.1 / ( 2.5 + 3.7 ) = 2.6359, but the answer is 9.56.

More step by step of my work:

Partial derivative of x is A2.5cos(2.5x+3.7y-43.1t)

This trend continues will all the other partial derivatives with A and cos(2.5x+3.7y-43.1t) being canceled out in the end. This would mean 2.5 is left for x, 3.7 is left for y, and -43.1 is left for t. Phi will also cancel. Now

0=2.5+3.7-43.1/v^2

v=sqrt(43.1/(2.5 + 3.7))

This gives the wrong answer of 2.64.

the speed here would be the square root of the partial t derivative divided by the sum of the partial x and y derivatives.

the amplitude, phi and the cos portion of the partial derivatives would all cancel out.

What I am left with is the sqrt(43.1 / ( 2.5 + 3.7 ) = 2.6359, but the answer is 9.56.

More step by step of my work:

Partial derivative of x is A2.5cos(2.5x+3.7y-43.1t)

This trend continues will all the other partial derivatives with A and cos(2.5x+3.7y-43.1t) being canceled out in the end. This would mean 2.5 is left for x, 3.7 is left for y, and -43.1 is left for t. Phi will also cancel. Now

0=2.5+3.7-43.1/v^2

v=sqrt(43.1/(2.5 + 3.7))

This gives the wrong answer of 2.64.