tangent line eq from parametric equation to rectangulat

[quazzimotto]

Find tangent line equation @ point t=1 for:
t=1
x=3t^2-2t+1
y=1/t

when t=1, x=2,y=1 from above equations: I get answer of y = 1-(1/4)x as detailed in link. Does this appear corrrect?


http://home.comcast.net/~0server0/tan.jpg

thanks

 

[pka]

Can you show how you got that answer?

 

[soroban]

Hello, quazzimotto1

An arithmetic error . . .

Find tangent line equation @ point $\displaystyle t=1$ for: $\displaystyle x\:=\:3t^2\,-\,2t\,+\,1,\;\;y\:=\:\frac{1}{t}$

When $\displaystyle t=1,\:x=2,\:y=1$: I get: $\displaystyle \:y\:=\:1\,-\,\frac{1}{4}x\;$ . . . Sorry, no

Your slope is correct . . . $\displaystyle \frac{dy}{dx}\,=\,-\frac{1}{4}$

Then we have: $\displaystyle \:y\,-\,1\:=\:-\frac{1}{4}(x\,-\,2)$

Hence: \(\displaystyle \:y\,-\,1\:=\:-\frac{1}{4}x\,+\,\frac{1}{2}\;\;\Rightarrow\;\;\L y\:=\:\frac{3}{2}\,-\frac{1}{4}x\)

 

[quazzimotto]

I'm just a little confused how the equation sets up:

So in y-1, the 1 comes from original y value @ t=0=1, x=2, y=1
and in the x-2, the 2 comes from the original X value from t=0=1, x=2, y=1?
thanks

Hello, quazzimotto1

An arithmetic error . . .

Find tangent line equation @ point $\displaystyle t=1$ for: $\displaystyle x\:=\:3t^2\,-\,2t\,+\,1,\;\;y\:=\:\frac{1}{t}$

When $\displaystyle t=1,\:x=2,\:y=1$: I get: $\displaystyle \:y\:=\:1\,-\,\frac{1}{4}x\;$ . . . Sorry, no

Your slope is correct . . . $\displaystyle \frac{dy}{dx}\,=\,-\frac{1}{4}$

Then we have: $\displaystyle \:y\,-\,1\:=\:-\frac{1}{4}(x\,-\,2)$

Hence: \(\displaystyle \:y\,-\,1\:=\:-\frac{1}{4}x\,+\,\frac{1}{2}\;\;\Rightarrow\;\;\L y\:=\:\frac{3}{2}\,-\frac{1}{4}x\)

 

[stapel]

I get answer of y = 1-(1/4)x as detailed in link.

Your image is a 2.7 meg graphic that measures 5102 by 6599 pixels and displays almost entirely textual material.

In the future, it might be helpful if you posted content that most of us could reasonably view. I have surfed, attempted to view, downloaded, and re-opened the graphic, in order to see what you did. Most won't bother.

Thank you for your consideration.

Eliz.

======================================================

P.S. to other viewers: This is what was in the graphic:

x² + y² = r². . .$\displaystyle y\,=\,\pm\,\sqrt{r^2\,-\,y^2}$

6) Find the equation of the tangent line to the curve at the point t = 1. [student note: (2, 1)] The curve is defined by the following set of parametric equations:

. . . . .x = 3t² - 2t - 1
. . . . .y = 1 / t

. . . . .t. . . . .x. . . . .y
. . . . .-------------------------
. . . . .0. . . . .1. . . . .undef
. . . . .1. . . . .2. . . . .1

[student note: graph looks like positive half of y = 1/x, with directional arrows leading down and then to the right]

1) Find derivative to get slope of tangent line

. . . . .\(\displaystyle \L \frac{dy}{dx}=\frac{\left(\frac{dy}{dt}\right)}{\left(\frac{dx}{dt}\right)}=\frac{(-1)t^{-1-1}}{6t-2}=\frac{\left(\frac{-1}{t^2}\right)}{\left(\frac{6t-2}{1}\right)}=\frac{-1}{t^2(6t-2)}=\frac{-1}{2t(3t-1)}\)

When t = 1, then m = -1/4.

2) Derivative in terms of t which is slope needs to be converted to x,y coordinate pair.

When t = 1, x = 2, y = 1.

. . . . .x. . . . . . . .|. . . . .y
---------------------|------------------
2 = 3t² - 2t + 1. .|. . . . .1 = 1/t
1 = t(3t - 2). . . ..|
solve (). . . . . . . .|
t = -1/3 or t = 1.|. . . . .t = 1

Any t which appears in both lists will give point.

. . . . .\(\displaystyle \L \frac{dy}{dx}=\frac{-1}{2t^2(3t-1)}=\frac{-1}{2(1)^2(3t-1)}=\frac{-1}{4}\)

Tangent line is given by Y = F(x), where x = a
Then Y = F(a) + (dy/dx)(x - a), F(a) = x = 2

Since, at t = 1, x = 2, y = 1, and since m = dy/dx, then
Tangent Line Equation --> Y = 1 - (1/4)x

 

[quazzimotto]

resized jpg ... , did not know it was so large.

Wittled it down , thanks for letting me know