Consider the curve created by&nbsp;2x2+3y2–4xy=36(a) Show that&nbsp;dydx=2y−2x3y−2x(b) Calculate the slope of the line perpendicular to the curve at each location where x is.&nbsp;x=6.(c) A vertical tangent line to the curve exists at a positive value of x.&nbsp;

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Consider the curve created by&amp;nbsp;2x2+3y2–4xy=36(a) Show that&amp;nbsp;dydx=2y−2x3y−2x(b) Calculate the slope of the line perpendicular to the curve at each location where x is.&amp;nbsp;x=6.(c) A vertical tangent line to the curve exists at a positive value of x.&amp;nbsp;

usumbiix · Accepted Answer

So,2x2+3y2−4xy=36&amp;nbsp;d&amp;nbsp;dx&amp;nbsp;(2x2+3y2−4xy)=d&amp;nbsp;dx&amp;nbsp;(36)&amp;nbsp;d&amp;nbsp;dx&amp;nbsp;(2x2)+d&amp;nbsp;dx&amp;nbsp;(3y2)−d&amp;nbsp;dx&amp;nbsp;(4xy)=0&amp;nbsp;let&amp;nbsp;u=3y2&amp;nbsp;du&amp;nbsp;dx&amp;nbsp;=du&amp;nbsp;dy&amp;nbsp;⋅&amp;nbsp;dy&amp;nbsp;&amp;nbsp;dx&amp;nbsp;=d(3y2y)⋅&amp;nbsp;dy&amp;nbsp;&amp;nbsp;dx&amp;nbsp;=6y⋅&amp;nbsp;dy&amp;nbsp;&amp;nbsp;dx&amp;nbsp;&amp;nbsp;4x+6y⋅&amp;nbsp;dy&amp;nbsp;&amp;nbsp;dx&amp;nbsp;−(4x⋅&amp;nbsp;dy&amp;nbsp;&amp;nbsp;dx&amp;nbsp;+4y)=0&amp;nbsp;4x+6y⋅&amp;nbsp;dy&amp;nbsp;&amp;nbsp;dx&amp;nbsp;−4x⋅&amp;nbsp;dy&amp;nbsp;&amp;nbsp;dx&amp;nbsp;−4y=0&amp;nbsp;&amp;nbsp;dy&amp;nbsp;&amp;nbsp;dx&amp;nbsp;(6y−4x)+4x−4y=0&amp;nbsp;&amp;nbsp;dy&amp;nbsp;&amp;nbsp;dx&amp;nbsp;(6y−4x)=4y−4x&amp;nbsp;&amp;nbsp;dy&amp;nbsp;&amp;nbsp;dx&amp;nbsp;=4y−4x6y−4x=2(2y−2x)2(3y−2x)&amp;nbsp;&amp;nbsp;dy&amp;nbsp;&amp;nbsp;dx&amp;nbsp;=2y−2x3y−2x

Corgnatiui · Accepted Answer

Part B : slope of the line tangent to the curve at each point on the curve where x=6.  The slope of the curve will be dydx=2y−2x3y−2x  we need to find slope of line tangent to curve at x=6  first we need find y value at x=6. For that substitute x=6 in equation of curve.  2x2+3y2−4xy=36  2(62)+3y2−4×6×y=36  72+3y2−24y−36=0  3y2−24y+36=0  divide by 3 we get,  y2−8y+12=0  (y−6)(y−2)=0  therefore y=6   or   y=2  therefore the point off tangency are   (x,y)=(6,6)       and     (x,y)=(6,2)  equation of slope =dydx=2y−2x3y−2x=0  at (x,y)=(6,6)    slope m1=2×6−2×63×6−2×6=0  m1=0  (x,y)=(6,2)  slope m2=2×2−2×63×2−2×6=4−126−12=86=43  m2=43

karton · Accepted Answer

Part C : value of x at which the curve has a vertical tangent line The vertical tangent to a curve occurs at a point where slope is undefined or infinite. This is nothing but the points where derivative of function is not defined. we know the derivative of function is, dydx=2y−2x3y−2x dydx is undefined at points where the denominator becomes zero. that is 3y − 2x = 0 3y = 2x y=23x substitute this in equation of curve, 2x2+3y2−4xy=362x2+3(23x)2−4x(23x)=362x2+43x2−83x2=362x2−43x2=36multiply throughiut by 3,6x2−4x2=1082x2=108x2=1082=54x=±54=±7.348 therefore vertical tangent to the curve occurs at points x = +7.348 and x = - 7.348

Consider the curve defined by 2x^{2} + 3y^{2} – 4xy = 36Z

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