second drivitive of a parametric

I have a parametrically defined curve where i am trying to find the second derivative.

here is the problem:

$\displaystyle x(t) = \frac {1}{t}$

and

$\displaystyle y(t) = -2 + lnt $

I need to find the equation for the tangent line to the curve at the point defined by t = 1

finding $\displaystyle \frac{dy}{dx}$

the dirivitive of x = $\displaystyle -\frac{1}{t^2}$

the dirivitive of y = $\displaystyle \frac{1}{t}$

then, i evaluate t = 1 by plugging it into

Sorry having some formatting issues below i hope you can read it..

$\displaystyle \frac{dy}{dt} \frac{1}{t}$

_______

$\displaystyle \frac{dx}{dt} \frac{-1}{t^2}$

after plugging in 1 for t

I get a slope of -1

so, then i need to take a second derivative

$\displaystyle \frac{d^2y}{dx^2} $

for the numerator its just 1

for the denominator its the derivative of x = $\displaystyle -\frac{1}{t^2}$

if so how do i evaluate that?

Re: second drivitive of a parametric

Quote:

Originally Posted by

**icelated** I have a parametrically defined curve where i am trying to find the second derivative.

here is the problem:

$\displaystyle x(t) = \frac {1}{t}$

and

$\displaystyle y(t) = -2 + lnt $

I need to find the equation for the tangent line to the curve at the point defined by t = 1

finding $\displaystyle \frac{dy}{dx}$

the dirivitive of x = $\displaystyle -\frac{1}{t^2}$

the dirivitive of y = $\displaystyle \frac{1}{t}$

then, i evaluate t = 1 by plugging it into

Sorry having some formatting issues below i hope you can read it..

$\displaystyle \frac{dy}{dt} \frac{1}{t}$

_______

$\displaystyle \frac{dx}{dt} \frac{-1}{t^2}$

after plugging in 1 for t

I get a slope of -1

so, then i need to take a second derivative

$\displaystyle \frac{d^2y}{dx^2} $

for the numerator its just 1

for the denominator its the derivative of x = $\displaystyle -\frac{1}{t^2}$

if so how do i evaluate that?

You have a first derivative of $\displaystyle \dfrac{dy(t)}{dx(t)} = -t$

To find the second derivative you need to differentiate the first derivative which should be quite simple

Re: second drivitive of a parametric

Quote:

Originally Posted by

**e^(i*pi)**

To find the second derivative you need to differentiate the first derivative which should be quite simple

Thats what i thought - so there isnt a second derivative! or rather zero

So im left with just

$\displaystyle \frac{-1}{t^2}$

for the second derivitive?

Re: second drivitive of a parametric

Quote:

Originally Posted by

**icelated** Thats what i thought - so there isnt a second derivative! or rather zero

So im left with just

$\displaystyle \frac{-1}{t^2}$

for the second derivitive?

no ... you need to take the derivative w/ respect to x

Re: second drivitive of a parametric

Quote:

Originally Posted by

**icelated** Thats what i thought - so there isnt a second derivative! or rather zero So im left with just

$\displaystyle \frac{-1}{t^2}$ for the second derivitive?

You have the wrong idea about the second derivative.

$\displaystyle \frac{{d^2 y}}{{dx^2 }} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{{\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)}}{{\frac{{dx}}{{dt}}}}$.

This may be simpler notation, is not strictly correct.

Let $\displaystyle y'=\frac{dy}{dx}$ then $\displaystyle \frac{d^2y}{dx^2}=\frac{dy'/dt}{dx/dt}$

Re: second drivitive of a parametric

Hi plato - oh, lets see if i got this right..

so i take the derivative of y prime

over the derivative of x

Y`` = $\displaystyle \frac{1}{t}$

because i dont need to take the 2nd derivative of denominator.

so i end up with the numerator as the same as the denominator?

Re: second drivitive of a parametric

Quote:

Originally Posted by

**icelated** Hi plato - oh, lets see if i got this right..

so i take the derivative of y prime

over the derivative of x

Y`` = $\displaystyle \frac{1}{t}$

because i dont need to take the 2nd derivative of denominator.

so i end up with the numerator as the same as the denominator?

The first derivative is:

$\displaystyle y'=\frac{dy/dt}{dx/dt}=-t$.

So $\displaystyle \frac{d^2y}{dx^2}=~?$

Re: second drivitive of a parametric

Well if $\displaystyle \frac{dy}{dx}$

= $\displaystyle \frac{1}{t}$

then, i take the second derivitive of that in respect to x but i cant do a

fraction over a fraction in latex.

taking the derivitive of $\displaystyle \frac{1}{t}$ in respect to x

I get $\displaystyle \frac{-1}{t^2}$

Which is exactly as the same as $\displaystyle \frac{dx}{dt}$

So they cancel out for

$\displaystyle \frac{d^2y}{dx^2}$ = 0

Right?

Re: second drivitive of a parametric

$\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}(-t) = -1 \cdot \frac{dt}{dx} = -1 \cdot (-t^2) = t^2$