Curves.

**richardsim10** · November 28th 2012, 03:36 PM

Ok, so I have this line equation and its graph:

Line D's equation is y = x. Now the first question asks me what is the point of intersection of C, defined as r(t) and D. That was easy, I just set x=t and y=t and solve each components seperately. I've found that they intersect at t=1, or at point (1,1).

Now the second question has two parts: Do any tangent vectors at C parallel to the line D? If so, find those points and if there are any tangents vectors perpendicular to line D, find those points too. If for either case it doesn't exist, prove it.
Now, how I did it was I took the derivative of r(t), then for the first part, set both components equal to t, which is the equation of the line. For the y component I,ve found t=3, but for x component I've found t=(1/4)*(+-1-sqrt(17)). I'm wondering about the validity of my ts'. For the second part, which asks for points at C perpendicular to D, I did the same thing except I equalled both components to -t,which I figured is perpendicular to D.
I'm doubting myself if my method is correct, can someone confirm if it is correct and if not, how I should approach it?

**skeeter** · November 28th 2012, 03:48 PM

if a tangent vector is parallel to $y = x$ , then ...

$\frac{dy}{dx} = 1 \implies \frac{dy}{dt} = \frac{dx}{dt}$

$2t-3 = 3t^2-2$

$0 = 3t^2-2t+1$

$b^2-4ac = 4 - 4(3)(1) < 0$

... no real solution, so no parallel tangent vectors

if a tangent vector is perpendicular to $y = x$ , then ...

$\frac{dy}{dx} = -1 \implies \frac{dy}{dt} = -\frac{dx}{dt}$

$2t-3 = 2 - 3t^2$

$3t^2 + 2t - 5 = 0$

$(3t + 5)(t - 1) = 0$

can you finish?

**richardsim10** · November 28th 2012, 03:52 PM

Yes, that's the thing I was missing. thank you very much.

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