Question

Find all points on the curve r(t)= (t, t^2,t^3) where its tangent line is parallel to the vector v = i +4j +12k .
Anyone help me, pleae

Answer 1

So if the tangent to that curve is parallel to the vector v, what can you say?

Call the tangent T. T is parallel to v is there is a constant k such that
T = kv,
yes?

Now, T changes with time T(t). You find it by differentiating r(t); i.e.,
T(t) = r'(t)

Now see what that implies for different values of t.

Answer 2

yes. I got \[T(t)=\frac{ <1,2t,3t^2> }{ \sqrt{1+4t^2+9t^4}? }\]

Answer 3

No need to normalize the T vector

Answer 4

and then break it down under the form of a vector. since vector v = <1,4,12> I let \[\frac{ 1 }{\sqrt{1+4t^2+9t^4} } = k*1(1 from x value of vector v)\]

Answer 5

yes. But again, loose the normalization, as it isn't necessary and it will significantly simplify your calculations.

Answer 6

and let Ty =k Vy
Tz =kVz
and solve the system. if you don't break the T, how to find out the k

Answer 7

please show me the stuff. I really stuck there

Answer 8

using the unnormalized tangent vector
<1,2t,3t^2> = k<1,4,12>

Hence
1 = k
2t = 4k
3t^2 = 12k

Now, for what values of t can this be true?

Answer 9

2

Answer 10

t = 2, yes. Hence there is only one point on the curve where the tangent vector is parallel to v, and that is r(t=2) = .....

Answer 11

can we ignore the denominator of T(t)? because that vector totally different from r(t)' which is just numerator of T(t)

Answer 12

All we are checking is whether two vectors are parallel. In general, if v and w are parallel,
then av and bw are also parallel for any non-zero constants a and b.

So here, we are simplifying the calculations by just using the unnormalized form of T.

Answer 13

in fact, v and w are parallel if and only if av and bw are parallel for any non-zero constants a and b.

Answer 14

Thank you very much. can you give me some thing related to "unnormalized" form as conference to read (some thing like theorem, article or else.) i want to make sure 100% understand what i am doing. anyway, thanks a lot