find unit vectors that are parallel to the tangent line to the parabola y=x^2 at the point (2,4).

Question

amistre64 · Answer

whats the derivative of the parabola?

anonymous · Answer

y'=2x, y'(2)=4

amistre64 · Answer

good, at x=2, the derivative = 4

what is the length of this vector? <2,4>

anonymous · Answer

(2,4) is a point, is you <2,4> from the derivative?

amistre64 · Answer

im used to vector notation being place int < , > wrappers

but we can also think of this as finding the distance from the point (2,4) from the origin

anonymous · Answer

but is the slope of the unit vectors to be found =4?

amistre64 · Answer

no

anonymous · Answer

length is sqrt(20), then <2/sqrt(20), 4/sqrt(20)> ?

amistre64 · Answer

correct, and simplify as wanted

anonymous · Answer

So the slope of the tangent line to y=x^2 at x=2 is not 4?

amistre64 · Answer

.... gonna make me have to reconsider my whole life arent you lol

anonymous · Answer

i see, i was thinking of beside it, instead of on it-, And how to you find more? let say move up from the first vector we found

amistre64 · Answer

yes, the slope at x=2 is 4 .... thanx

amistre64 · Answer

our vector is then <1,4> ; not <2,4>

amistre64 · Answer

what do you mean by move up from the vector?

anonymous · Answer

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