find parametric equations for the tangent line at t=2 for x=t^2, y=t^3

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find parametric equations for the tangent line at t=2 for x=t^2, y=t^3

Mathematics
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Okay, sounds like you need to find the tangent line of this curve at t=2 and state that tangent line in terms of some parameter. Give me a minute to write something out and scan, okay?
k
scanning

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k
I know it might look a bit full-on. Does that stuff look familiar?
yah i did the problem but there is a discrepancy bw my answer an the answer found at the end of the book
the answer i got was x=4+4t and y=8+12t but at the end of the book the answer was x=4+4(t-2) and y=8+12(t-2_ do u know why?
Well, just looking at that, I don't see any difference between what you have and what the book does, because if you identify t-2 as a parameter, what's being said between you guys is\[x=4+4(paramater)\]\[y=8+12(parameter)\]
If you were to take ALL the real numbers and plug them into t and t-2, you'd end up with the same set of numbers, and because the form of both your equations is the same, you'd end up with the same lines.
so there is no difference
no real difference
but do u know how it becomes t-2
Well, I had a look at my solution and their solution and they're identical, just parametrized differently. It's the case with parametrization that you can set x equal to ANY kind of concoction of some parameter you want, and because of the relationship between y and x, derive the associated parametrization for y. Here, I chose s, but they chose 4+4(t-2). My choosing s forced my y to be 3s-4, and their choosing x=4+4(t-2) forced their y to be 8 + 12(t-2). Why they chose that parametrization, I cannot say. They might be setting the problem up for something to come.
okay thank u for ur help
np
Incidentally, how did you get your own parametrizations?
u mean how did i find the answer
Well, how did you go about parametrizing? What did you say 'let x equal' to?
I guess I mean, what was your motivation for 4+4t?
well i first took the derivative , plugged in 2 into each equation to find my slope . then plugged in two in two into the original equations to find a point and plugged everything into the x and y formula
By x and y formula, do you mean the point-slope formula for a straight line?

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