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If π equals negative π’ minus π£, then the polar form of π is blank.
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(A) Root two, π over four, (B) root two, three π over four, (C) root two, five π over four, (D) root two, seven π over four.
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All right, so given this vector π in rectangular form, we want to solve for its polar form.
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Another way we can write π in rectangular form is to express it in terms of its π₯- and π¦-components like this.
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And now letβs recall that for a vector expressed in polar form, we give it not by its π₯-, π¦-components, but rather by π and π.
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Here, π is equal to the square root of π₯ squared plus π¦ squared, and the tan of π is equal to π¦ divided by π₯.
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Using these relationships, we then have the ability to convert π from its rectangular form as given to its polar form.
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That polar form, as weβve seen, is defined by a radial distance and an angle.
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And π, we know, is equal to the square root of the π₯-component squared plus the π¦-component squared.
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This is equal to the square root of one plus one or the square root of two, so weβll substitute this result in for π in our polar form of π.
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But as we look again at our answer options, notice that this doesnβt narrow down the list.
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All four choices had the same π-value of the square root of two.
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So letβs move on to calculating the angle π of our vector.
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As we do this, it will be helpful to sketch our vector on a coordinate plane.
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Letβs say that each of these tick marks represents one unit of distance.
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And since our vector has rectangular components of negative one and negative one, if the tail of the vector was at the origin, then the vector would look like this.
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And the angle π, defining its direction, would be measured from the positive π₯-axis to the vector.
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We see then that π will be greater than π radians but less than three-halves π.
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We can now use this relationship here to solve for it.
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We know that the components of our vector π₯ and π¦ are both negative one.
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Here weβve taken the inverse tan of both sides of our tan of π equation, meaning that π equals the arc or inverse tan of negative one divided by negative one.
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Hereβs whatβs interesting, though.
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If we evaluate this inverse tangent on our calculator, the result we get is exactly π divided by four.
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But looking at our sketch of our vector, we know that that canβt be the correct angle for π.
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Here, we need to recall the rule that when we calculate an inverse tangent with a negative π₯-value in the fraction, then to correctly solve for the angle π measured relative to the positive π₯-axis, we need to add π radians to our result.
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Itβs by doing this that we avoid a possible error in calculating π.
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This potential error can be traced back to the tangent function.
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But suffice to say whenever we calculate the inverse tangent of a fraction with a negative π₯-value, that is, whenever our vectorβs in the second or third quadrants, weβll need to add π radians or 180 degrees, as the case may be, to properly solve for the angle relative to the positive π₯-axis.
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π over four plus π is equal to five over four times π.
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And substituting this value in for π in our polar form of π, we see that in polar form π is equal to the square root of two, five π over four.
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We find that listed as option (C) among our answer choices.
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So, to complete our sentence, if π equals negative π’ minus π£, then the polar form of π is square root of two, five π over four.