# Thread: second drivitive of a parametric

1. ## 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?

2. ## Re: second drivitive of a parametric

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

3. ## Re: second drivitive of a parametric

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?

4. ## Re: second drivitive of a parametric

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

5. ## Re: second drivitive of a parametric

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

6. ## 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?

7. ## Re: second drivitive of a parametric

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

8. ## 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?

9. ## Re: second drivitive of a parametric

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