May 14th, 2016, 06:34 AM  #1 
Newbie Joined: May 2016 From: england Posts: 7 Thanks: 0  differentiation question
hi im given $\displaystyle x=t^3$and$\displaystyle Y=t^24t$ then im asked to find $\displaystyle dy/dx$ in terms of t, find the slope of this curve at (1,6) and then determine a point at which the tangent to this curve is parallel to the xaxis. i worked out $\displaystyle (dy/dx) = (2t^24)/(3t^2)$ second part i used the point (1,6) to a slope of 2 3rd part i dont know how to do, correct me if i went wrong anywhere 
May 14th, 2016, 07:19 AM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 3,094 Thanks: 1677 
$x=t^3 \implies \dfrac{dx}{dt}=3t^2$ $y=t^24t \implies \dfrac{dy}{dt}=2t4$ $\dfrac{dy}{dx}=\dfrac{2t4}{3t^2}$ a tangent line parallel to the xaxis would have a slope equal to zero, correct? What value of $t$ would make $\dfrac{dy}{dx}=0$ ? 
May 14th, 2016, 07:26 AM  #3 
Newbie Joined: May 2016 From: england Posts: 7 Thanks: 0 
t = 2 ? $\displaystyle 2(2)4/3(2^3) $ $\displaystyle 0/24 = 0 $ 
May 14th, 2016, 07:59 AM  #4 
Math Team Joined: Jul 2011 From: Texas Posts: 3,094 Thanks: 1677  
May 14th, 2016, 08:06 AM  #5 
Newbie Joined: May 2016 From: england Posts: 7 Thanks: 0 
(0,2) not sure do i let my dy/dt = 0 to get my y point and dx/dt = 0 to get my x point?
Last edited by rath; May 14th, 2016 at 08:10 AM. 
May 14th, 2016, 08:25 AM  #6 
Math Team Joined: Jul 2011 From: Texas Posts: 3,094 Thanks: 1677 
the point is on the curve, $\left(x(2),y(2) \right)$ ...


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differentiation, line, question, slope 
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