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

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay, 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?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I know it might look a bit fullon. Does that stuff look familiar?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yah i did the problem but there is a discrepancy bw my answer an the answer found at the end of the book

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the answer i got was x=4+4t and y=8+12t but at the end of the book the answer was x=4+4(t2) and y=8+12(t2_ do u know why?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, just looking at that, I don't see any difference between what you have and what the book does, because if you identify t2 as a parameter, what's being said between you guys is\[x=4+4(paramater)\]\[y=8+12(parameter)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you were to take ALL the real numbers and plug them into t and t2, 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so there is no difference

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but do u know how it becomes t2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, 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(t2). My choosing s forced my y to be 3s4, and their choosing x=4+4(t2) forced their y to be 8 + 12(t2). Why they chose that parametrization, I cannot say. They might be setting the problem up for something to come.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay thank u for ur help

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Incidentally, how did you get your own parametrizations?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u mean how did i find the answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Well, how did you go about parametrizing? What did you say 'let x equal' to?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I guess I mean, what was your motivation for 4+4t?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0By x and y formula, do you mean the pointslope formula for a straight line?
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