(a) Cartesian to Polar Take the square root of both sides of the equation to fin

Question

(a) Cartesian to Polar

Take the square root of both sides of the equation to find the radial coordinate.

Use the tangent function to find the angle with the inverse tangent, adding 180° because the angle is actually in third quadrant.

r = x2 + y2 = (-3.50)2 + (-2.50 m)2 = 4.30 m

= tan-1 (0.714) = 35.5° + 180° = 216°

(b) Polar to Cartesian

Use the trigonometric definition equations.

x = r cos = (5.00 m) cos 37.0° = 3.99 m

y = r sin = (5.00 m) sin 37.0° = 3.01 m

Remarks When we take up vectors in two dimensions, we will routinely use a similar process to find the direction and magnitude of a given vector from its components, or, conversely, to find the components from the vector's magnitude and direction.

Question If you start with the answers in part (b) and work backwards to recover the radius and angle, there will be slight differences from the original quantities. What caused this difference? (Select all that apply.)

A) Using inconsistent equations in doing the calculation in both directions.

B) Calculator defects.

C) Rounding the value of the tangent of the angle before taking its cosine in the example.

D) Rounding the final calculated values of x and y in the example before using them to work backwards.

E) Keeping more than three significant figures in intermediate steps of each calculation.

The equations in both directions are equivalent. Examine the places in the example where calculated values are rounded before using them in the next step. Consider what effect that has in introducing uncertainties in the calculation? Would smaller or larger discrepancies result if you keep a lot of extra digits in the calculation until the final answer?

Explanation / Answer

Following are the sources of error

B) Calculator defects.

C) Rounding the value of the tangent of the angle before taking its cosine in the example.

D) Rounding the final calculated values of x and y in the example before using them to work backwards.

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