Calculus
posted by Sammy .
For the parametric curve defined x=2t^312t^230t+9 and y=t^24t+6 ..find dy/dx and then where is the tangent to the curve vertical (give the cartesian coordiantes of the points. and find the tang. to the curve horizontal
dx/dt= 6t^224t30
dy/dt= 2t4
dy/dx= (t2)/(3t^212t15)
Now, if dy/dx= inf, then
3t^212t15=0 or
(t5)(t+1)=0 check that.
solve for t. Put that value of t in the original equations to get x,y for the point where the tangent is vertical.
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