Question

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).

Answer 1

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.

Answer 2

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.

Answer 3

Your dy/dx is\[\frac{ dy }{ dx }=\frac{ \frac{ dy }{ dt } }{ dx/dt }\]

Answer 4

I'd guess you already know how to find dy/dt and dx/dt.

Answer 5

x'(t)=4, y'(t)=4t^3.

Answer 6

WOAH wait wrong set

Answer 7

You'll have to take the given point (1,6) and work backward to find the t value that gives you x and y.

Answer 8

x'(t)=-2/x^2, y'(t)=2t

Answer 9

If x(t)=t/2, find dx/dt. It's not 4.
If y(t)=t^2+2, find dy/dt.

Answer 10

My bad, it's x(t)=2/t.

Answer 11

Well, if x(t)=2/t, then x(t) = 2*(1/t) or 2*t^(-1). Find dx/dt.

Answer 12

I'd use the power rule, were I you.

Answer 13

@RightInTheCranium: sorry, but I have to ask you to hurry. Bedtime for me! ;)

Answer 14

Er...x'(t)=-2/t^2 and y'(t)=2t isn't right?

Answer 15

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.

Answer 16

\[\frac{ 2t }{-2/t^2 }\]?

Answer 17

\[\frac{ dy }{ dx }=\frac{ \frac{ dy }{ dt } }{ \frac{ dx }{ dt } }=?\]

Answer 18

Good. Now reduce that result of yours.

Answer 19

Bleh, typing in this site can be so slow sometimes. It's -t^3.

Answer 20

Yes. Now we have to find the value of t so that we can find the value of this derivative.

Answer 21

We're given a point on the curve; it is (1,6).

what is the formula for x(t)? Type it here.

Answer 22

x(t)=2/t

Answer 23

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.

Answer 24

t = ?

Answer 25

t=2. So -t^3=-8 ?

Answer 26

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.

Answer 27

y-6=-8(x-1) => y-8x+2=0 ?

Answer 28

Or wait

Answer 29

y-8x-14=0

Answer 30

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.

Answer 31

Your y-6=-8(x-1) is correct. You could put it into a fancier form later if you wish.

Answer 32

Hmm...I'm looking at the answer choices I have for this questions and none of them look like what I got.

Answer 33

I'd suggest you double check to ensure that you have your parametric equations for x and y correct.

Answer 34

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!

Answer 35

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.

Answer 36

what are your answer choices @RightInTheCranium ?

Answer 37

Signing off now. Good night. Jim: Thanks for your input!

Answer 38

\[2x+\frac{ 5 }{ 4 }\] \[4x-\frac{ 11 }{ 2 }+\frac{ y }{ 4 }\] \[2x-\frac{ 4 }{ 4 }\] \[4x-7+\frac{ y }{ 2 }\] -4x+3

Answer 39

hmm that's strange, none of them are equivalent to y = -8x+14. I was hoping at least one was

Answer 40

Hmm...I solved for t from the x(t) equation. Maybe the y(t) will yield something.

Answer 41

Hm, gonna keep trying. Setting t^2+2=6 might give me something. Hopefully?

Answer 42

I already posted that above in a previous post

Answer 43

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.)

Answer 44

I got the last option
here is why i would choose that