# Homework Help: Find direction of acceleration from potential-energy funct.

1. Oct 18, 2016

### Grawlix

1. The problem statement, all variables and given/known data
A small block with mass 0.0400 kg is moving in the xy-plane. The net force on the block is described by the potential-energy function U(x, y) = (5.80 J/m2)x2-(3.60 J/m3)y3. What are the magnitude and direction of the acceleration of the block when it is at the point (x = 0.300 m, y = 0.600 m)?

2. Relevant equations
U(x, y) = (5.80 J/m2)x2-(3.60 J/m3)y3

3. The attempt at a solution
I've attached the solution as an image; it was given to me with the question. Why does the x-component of acceleration becomes positive when plugged into arctan to find the angle?

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• ###### Why is x component of acceleration positive..PNG
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2. Oct 18, 2016

### Staff: Mentor

Hi Grawlix,

The arctan function takes one argument. We typically concoct that argument as a ratio of two numbers, the y and x components of a vector. So something like: θ = arctan(y/x).

Now the problem that arises is that if one or the other of the x or y components is negative, the arctan function cannot tell whether to assign the negative sign to the numerator or denominator as it only "sees" the result of the division y/x. Similarly, if both terms are negative the arctan function sees a positive argument, and there's no indication that either term was ever negative.

So it falls to the user of the arctan function to correctly place the resulting angle in the correct quadrant of the Cartesian plane according to his special knowledge of the signs of the components involved. This means adjusting the result of the arctan function accordingly.