find where dy/dx = 0
4 (x^2 + y^2)(2x + 2 y dy/dx) = 49 (2 x - 2 y dy/dx)
when dy/dx = 0
4(x^2+y^2)(2x) = 49(2x)
x^2+y^2 = 49/4 = r^2 it is a circle
center at (0,0) , r = 7/2
Find the points on the lemniscate, given below, where the tangent is horizontal
2(x^2 + y^2)^2 = 49(x^2 − y^2)
5 answers
i need four points how to find?
well, x^2 +y^2 = 49/4
put that back in
2 (49/4)^2 = 49 (x^2-y^2)
49/8 = x^2 - y^2
49/4 = x^2 + y^2
2 x^2 = 3*49/8
well that is two points x values :)
you have 2 y values for each x so 4
put that back in
2 (49/4)^2 = 49 (x^2-y^2)
49/8 = x^2 - y^2
49/4 = x^2 + y^2
2 x^2 = 3*49/8
well that is two points x values :)
you have 2 y values for each x so 4
thanks
Find the points on the lemniscate where the tangent is horizontal. (Order your answers from smallest to largest x, then from smallest to largest y.)
2(x2 + y2)2 = 49(x2 − y2)
2(x2 + y2)2 = 49(x2 − y2)