 My Math Forum curve parametric form differentiation help

 Calculus Calculus Math Forum

 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. Thanks from daveyf 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.) Thanks from daveyf 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 ... $y-\dfrac{1}{2}=\dfrac{1}{8}(x-6)$ $\dfrac{d}{dx} \left(\dfrac{1}{8}t^{-1} \right)$ $-\dfrac{1}{8}t^{-2} \cdot \dfrac{dt}{dx}$ $\left(-\dfrac{1}{8t^2} \right) \cdot \left(-\dfrac{t^3}{12} \right) = \dfrac{t}{96}$ Thanks from daveyf Tags curve, differentiation, form, parametric Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Mr Davis 97 Calculus 1 March 9th, 2015 05:50 AM Joselynn Calculus 5 September 9th, 2013 09:01 AM aaron-math Calculus 1 December 12th, 2011 05:59 PM dynezidane Calculus 2 June 1st, 2011 06:08 PM jared53 Calculus 1 November 30th, 2008 06:15 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      