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May 13th, 2016, 03:34 AM  #1 
Member Joined: Apr 2011 Posts: 46 Thanks: 0  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) $ 
May 13th, 2016, 04:48 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,697 Thanks: 2681 Math Focus: Mainly analysis and algebra 
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. 
May 13th, 2016, 05:11 AM  #3 
Member Joined: Apr 2011 Posts: 46 Thanks: 0 
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 
May 13th, 2016, 05:40 AM  #4 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
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.)

May 13th, 2016, 05:57 AM  #5 
Member Joined: Apr 2011 Posts: 46 Thanks: 0 
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. 
May 13th, 2016, 07:41 AM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 3,094 Thanks: 1677  $x=6t^{2} \implies \dfrac{dx}{dt}=12t^{3}=\dfrac{12}{t^3}$ $y=\dfrac{1}{2}t^{3} \implies \dfrac{dy}{dt}=\dfrac{3}{2}t^{4}=\dfrac{3}{2t^4}$ $\dfrac{dy}{dx}=\left(\dfrac{3}{2t^4}\right) \cdot\left(\dfrac{t^3}{12}\right)= \dfrac{1}{8t}$ $t=1$ at the given point ... $y\dfrac{1}{2}=\dfrac{1}{8}(x6)$ $\dfrac{d}{dx} \left(\dfrac{1}{8}t^{1} \right)$ $\dfrac{1}{8}t^{2} \cdot \dfrac{dt}{dx}$ $ \left(\dfrac{1}{8t^2} \right) \cdot \left(\dfrac{t^3}{12} \right) = \dfrac{t}{96}$ 

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curve, differentiation, form, parametric 
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