second drivitive of a parametric

**icelated** · December 8th 2011, 12:18 PM

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

here is the problem:

$x(t) = \frac {1}{t}$
and
$y(t) = -2 + lnt$

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

finding $\frac{dy}{dx}$

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

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

then, i evaluate t = 1 by plugging it into

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

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

after plugging in 1 for t

I get a slope of -1

so, then i need to take a second derivative
$\frac{d^2y}{dx^2}$

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

if so how do i evaluate that?

**e^(i*pi)** · December 8th 2011, 12:36 PM

Originally Posted by icelated

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

here is the problem:

$x(t) = \frac {1}{t}$
and
$y(t) = -2 + lnt$

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

finding $\frac{dy}{dx}$

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

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

then, i evaluate t = 1 by plugging it into

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

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

after plugging in 1 for t

I get a slope of -1

so, then i need to take a second derivative
$\frac{d^2y}{dx^2}$

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

if so how do i evaluate that?

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

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

**icelated** · December 8th 2011, 12:45 PM

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

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

for the second derivitive?

**skeeter** · December 8th 2011, 01:39 PM

Originally Posted by icelated

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

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

for the second derivitive?

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

**Plato** · December 8th 2011, 01:47 PM

Originally Posted by icelated

Thats what i thought - so there isnt a second derivative! or rather zero So im left with just
$\frac{-1}{t^2}$ for the second derivitive?

You have the wrong idea about the second derivative.
$\frac{{d^2 y}}{{dx^2 }} = \frac{d}{{dx}}\left( {\frac{{dy}}{{dx}}} \right) = \frac{{\frac{d}{{dt}}\left( {\frac{{dy}}{{dx}}} \right)}}{{\frac{{dx}}{{dt}}}}$ .

This may be simpler notation, is not strictly correct.

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

**icelated** · December 8th 2011, 01:58 PM

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

so i take the derivative of y prime
over the derivative of x

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

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

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

**Plato** · December 8th 2011, 02:04 PM

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`` = $\frac{1}{t}$
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:
$y'=\frac{dy/dt}{dx/dt}=-t$ .

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

**icelated** · December 8th 2011, 02:15 PM

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

= $\frac{1}{t}$

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 $\frac{1}{t}$ in respect to x

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

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

So they cancel out for

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

Right?

**skeeter** · December 8th 2011, 02:32 PM

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

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