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May 13th, 2016, 03:34 AM   #1
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curve parametric form differentiation help

hi need a little help with a part of this question.

a curve is given in parametric form by $\displaystyle X= 6/t^2 $ and $\displaystyle Y=1/2t^3$
i. find dy/dx in terms of t, which i worked out to be:
$\displaystyle dy/dx = (dy/dt)/(dx/dt) = (-12/t^3)/(-6/t^4)$

ii.find the equation of the tangent line to this curve at (6, 0.5)

not sure about this one i got$\displaystyle y = (19/36) - x/216 $

iii. find$\displaystyle ((d^2)y)/(dx^2)$ in terms of t: not sure about this either
i differentiate the answer to part one to get this using $\displaystyle ((d/dt)(dy/dx))/(dx/xt) $
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May 13th, 2016, 04:48 AM   #2
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Your derivative ${\mathrn d y \over \mathrm d x}$ looks incorrect to me. You have transposed the terms and the coefficient in $\mathrm d y \over \mathrm d t$ is incorrect. I don't think you misunderstand, they just look like mistakes.

For part ii), you will need to determine the value of $t$ that corresponds to the point $(x,y)=(6,\frac12)$. If you put that value of $t$ into your expression for $\mathrm d y \over \mathrm d x$ to find the slope of the curve at that point. You then have a point and the slope, and can therefore use the point/slope equation of a straight line to build the equation of the tangent.
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May 13th, 2016, 05:11 AM   #3
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so quotient rule seems the way to go for this? using it i got $\displaystyle (6t^2)/(2t^3) for dy/dt $ $\displaystyle (12t)/(t^2) for dx/dt $ and the slope was 6/2 for part 2

Last edited by daveyf; May 13th, 2016 at 06:10 AM. Reason: fixed mistake
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May 13th, 2016, 05:40 AM   #4
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Rather than use the quotient rule, it is simpler to write $\displaystyle y= (1/2)t^{-3}$ so that $\displaystyle y'= -(3/2)t^{-4}$. (Unless the function really $\displaystyle y= (1/2)t^3$. Writing $\displaystyle y= 1/2t^3$ it is unclear whether the $\displaystyle t^3$ is in the denominator or not.)
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May 13th, 2016, 05:57 AM   #5
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im not sure, this i uploaded the question here is Imgur: The most awesome images on the Internet but i think i went wrong the 1st time when i brought $\displaystyle Y=1/2t^3$ the 2 above the line with $\displaystyle t^3$ aswell to get $\displaystyle 1*2t^-3$

Last edited by daveyf; May 13th, 2016 at 06:03 AM.
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May 13th, 2016, 07:41 AM   #6
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$x=6t^{-2} \implies \dfrac{dx}{dt}=-12t^{-3}=-\dfrac{12}{t^3}$

$y=\dfrac{1}{2}t^{-3} \implies \dfrac{dy}{dt}=-\dfrac{3}{2}t^{-4}=-\dfrac{3}{2t^4}$

$\dfrac{dy}{dx}=\left(-\dfrac{3}{2t^4}\right) \cdot\left(-\dfrac{t^3}{12}\right)= \dfrac{1}{8t}$


$t=1$ at the given point ...

$y-\dfrac{1}{2}=\dfrac{1}{8}(x-6)$


$\dfrac{d}{dx} \left(\dfrac{1}{8}t^{-1} \right)$

$-\dfrac{1}{8}t^{-2} \cdot \dfrac{dt}{dx}$

$ \left(-\dfrac{1}{8t^2} \right) \cdot \left(-\dfrac{t^3}{12} \right) = \dfrac{t}{96}$
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