Posted by **Sammy** on Saturday, April 14, 2007 at 2:18pm.

For the parametric curve defined x=2t^3-12t^2-30t+9 and y=t^2-4t+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^2-24t-30

dy/dt= 2t-4

dy/dx= (t-2)/(3t^2-12t-15)

Now, if dy/dx= inf, then

3t^2-12t-15=0 or

(t-5)(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.

## Answer this Question

## Related Questions

- Calc. - sketch the curve using the parametric equation to plot the points. use ...
- calc - for the parametric curve defined by x=3-2t^2 and y=5-2t ...sketch the ...
- calculus - Given the curve defined by the equation y=cos^2(x) + sqrt(2)* sin(x) ...
- Calculus - Please help this is due tomorrow and I dont know how to Ive missed a ...
- Calculus - A curve is defined by the parametric equations: x = t2 – t and y = t3...
- 12th Grade Calculus - 1. a.) Find an equation for the line perpendicular to the ...
- Calculus - Consider the curve y^2+xy+x^2=15. What is dy/dx? Find the two points ...
- Calculus - the curve: (x)(y^2)-(x^3)(y)=6 (dy/dx)=(3(x^2)y-(y^2))/(2xy-(x^3)) a...
- calculus - 1. Given the curve a. Find an expression for the slope of the curve ...
- calculus - Consider the curve defined by 2y^3+6X^2(y)- 12x^2 +6y=1 . a. Show ...

More Related Questions