anonymous
  • anonymous
Find the equation in x and y for the line tangent to the curve x(t)=2/t, y(t)=t^2+2 at the point (1,6).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I know that you usually take the derivative of an x or y if you have it, but I don't know how to combine them both in a single equation.
mathmale
  • mathmale
You'll need to find the derivative, dy/dx, as usual. In this case y ou're working with "parametric equations," where t is your parameter.
mathmale
  • mathmale
Your dy/dx is\[\frac{ dy }{ dx }=\frac{ \frac{ dy }{ dt } }{ dx/dt }\]

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More answers

mathmale
  • mathmale
I'd guess you already know how to find dy/dt and dx/dt.
anonymous
  • anonymous
x'(t)=4, y'(t)=4t^3.
anonymous
  • anonymous
WOAH wait wrong set
mathmale
  • mathmale
You'll have to take the given point (1,6) and work backward to find the t value that gives you x and y.
anonymous
  • anonymous
x'(t)=-2/x^2, y'(t)=2t
mathmale
  • mathmale
If x(t)=t/2, find dx/dt. It's not 4. If y(t)=t^2+2, find dy/dt.
anonymous
  • anonymous
My bad, it's x(t)=2/t.
mathmale
  • mathmale
Well, if x(t)=2/t, then x(t) = 2*(1/t) or 2*t^(-1). Find dx/dt.
mathmale
  • mathmale
I'd use the power rule, were I you.
mathmale
  • mathmale
@RightInTheCranium: sorry, but I have to ask you to hurry. Bedtime for me! ;)
anonymous
  • anonymous
Er...x'(t)=-2/t^2 and y'(t)=2t isn't right?
mathmale
  • mathmale
If x(t)=2/t=2t^(-1), then x '(t) = -2/t^2, yes. Good. Now we're on the right track. Go back and look at my formula for finding dy/dx as a function of dy/dt and dx/dt. Write out your dy/dx for this particular problem.
anonymous
  • anonymous
\[\frac{ 2t }{-2/t^2 }\]?
mathmale
  • mathmale
\[\frac{ dy }{ dx }=\frac{ \frac{ dy }{ dt } }{ \frac{ dx }{ dt } }=?\]
mathmale
  • mathmale
Good. Now reduce that result of yours.
anonymous
  • anonymous
Bleh, typing in this site can be so slow sometimes. It's -t^3.
mathmale
  • mathmale
Yes. Now we have to find the value of t so that we can find the value of this derivative.
mathmale
  • mathmale
We're given a point on the curve; it is (1,6). what is the formula for x(t)? Type it here.
anonymous
  • anonymous
x(t)=2/t
mathmale
  • mathmale
Good. Now, the x value of the given point is 1. Set y our formula for x(t) equal to 1 and solve the resulting equation for t.
mathmale
  • mathmale
t = ?
anonymous
  • anonymous
t=2. So -t^3=-8 ?
mathmale
  • mathmale
Yes. That's the slope of the tangent line to the curve at (1,6). This tangent line obviously goes thru (1,6). Use your slope and this point to write the equation of the tangent line.
anonymous
  • anonymous
y-6=-8(x-1) => y-8x+2=0 ?
anonymous
  • anonymous
Or wait
anonymous
  • anonymous
y-8x-14=0
jim_thompson5910
  • jim_thompson5910
Alternative Route to find the slope: solve x = 2/t for t to get t = 2/x Then plug that into y = t^2 + 2 to get y = (4/x^2) + 2 now apply the derivative. I'm skipping a bunch of steps, but you'll end up with dy/dx = -8/x^3 Plugging in x = 1 leads to dy/dx = -8 The first method is more effective because often in parametric mode, it's hard to isolate t.
mathmale
  • mathmale
Your y-6=-8(x-1) is correct. You could put it into a fancier form later if you wish.
anonymous
  • anonymous
Hmm...I'm looking at the answer choices I have for this questions and none of them look like what I got.
mathmale
  • mathmale
I'd suggest you double check to ensure that you have your parametric equations for x and y correct.
anonymous
  • anonymous
Seems alright to me, but obviously there's something not quite right. Bloody hell. Oh well, time to ring up the professor. Thank you anyway!
mathmale
  • mathmale
Take care. Can't resist saying I hope you'll try this problem again on your own before seeing (or ringing up) your prof. We've covered a lot of ground and made a lot of progress.
jim_thompson5910
  • jim_thompson5910
what are your answer choices @RightInTheCranium ?
mathmale
  • mathmale
Signing off now. Good night. Jim: Thanks for your input!
anonymous
  • anonymous
\[2x+\frac{ 5 }{ 4 }\] \[4x-\frac{ 11 }{ 2 }+\frac{ y }{ 4 }\] \[2x-\frac{ 4 }{ 4 }\] \[4x-7+\frac{ y }{ 2 }\] -4x+3
jim_thompson5910
  • jim_thompson5910
hmm that's strange, none of them are equivalent to y = -8x+14. I was hoping at least one was
anonymous
  • anonymous
Hmm...I solved for t from the x(t) equation. Maybe the y(t) will yield something.
anonymous
  • anonymous
Hm, gonna keep trying. Setting t^2+2=6 might give me something. Hopefully?
jim_thompson5910
  • jim_thompson5910
I already posted that above in a previous post
1 Attachment
anonymous
  • anonymous
JUST when you think it's going to be easy. I'm don't know what I expected. Ugh. Time to consult the professor. Thanks anyway! (Again.)
triciaal
  • triciaal
I got the last option here is why i would choose that
triciaal
  • triciaal
|dw:1449931275110:dw|
triciaal
  • triciaal
|dw:1449931452362:dw|
triciaal
  • triciaal
|dw:1449931577907:dw|
triciaal
  • triciaal
|dw:1449931647813:dw|

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