Elements of analytical geometry and the differential and integral calculus . x and y represent the co-ordinates of the pointP, and X and y the general co-ordinates of the line, and a the tangent of its angle with the axis of X, then by (Prop. Ill,Chap. I,) we have y—y=a(x—x). Now the triangle APB gives us the following expression for the tangentof the angle APB, or its equal PTB, a=-^2 y This value of a put in the preceding equation, will give usy—y—~~~(x—x). y Or 2/2—yy=—x^-^xx. Whence yy-\-xx=R^ the same as before. Of the Polar Equation of the Circle. The polar equation of a curve is the equ
Elements of analytical geometry and the differential and integral calculus . x and y represent the co-ordinates of the pointP, and X and y the general co-ordinates of the line, and a the tangent of its angle with the axis of X, then by (Prop. Ill,Chap. I,) we have y—y=a(x—x). Now the triangle APB gives us the following expression for the tangentof the angle APB, or its equal PTB, a=-^2 y This value of a put in the preceding equation, will give usy—y—~~~(x—x). y Or 2/2—yy=—x^-^xx. Whence yy-\-xx=R^ the same as before. Of the Polar Equation of the Circle. The polar equation of a curve is the equation for any point inthe curve estimated from any fixed point called a pole. Thevariable distance from the pole to any point in the curve iscalled the radius vector, and the angle which the radius vectormakes with a given straight line is called the variable angle. PROPOSITION find the polar equation of the circle. When the center is the pole or the fixed point, the equation isx-+y^=^B^ (1) and the radius vector R is then constant. ANALYTICAL Now let P be the pole, and theco-ordinates of that point a and , and MPX=v the variableangle. ANz=x and NM=y. Thenit is obvious thatx-=a-\-r COS. v, and y±=J-(~**sin. values of x and y substi-tuted in (1), (observing that cos.^v-j-sin. ^a;=l,) will give which is the polar equation sought. Scholium 1. P may be at anypoint on the plane. Suppose it at^.Then a——R and ^=0. Substitu-ting these values in the equation,and it reduces to r^— cos. vr=^0. As there is no absolute term, r=0will satisfy one point in the curve,and this is true, as P is supposed to be in the curve. Dividingby r, and r=2i2cos. v. This value of r will be positive while cos. v is positive, andnegative when cos.^ is negative ; but r being a radius vector cannever be negative, and the figure shows this, as r never passesto the left of B, but runs into zero at that point. When 2;== 0, 008.^=1, then r=^^. When z^=90, cos,y=0,
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