My Math Forum differentiation question

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 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$\displaystyle Y=t^2-4t$ 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 x-axis. i worked out $\displaystyle (dy/dx) = (2t^2-4)/(3t^2)$ 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 $x=t^3 \implies \dfrac{dx}{dt}=3t^2$ $y=t^2-4t \implies \dfrac{dy}{dt}=2t-4$ $\dfrac{dy}{dx}=\dfrac{2t-4}{3t^2}$ a tangent line parallel to the x-axis 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

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Quote:
 Originally Posted by rath t = 2 ? $\displaystyle 2(2)-4/3(2^3)$ $\displaystyle 0/24 = 0$
ok ... now that we know the required value of $t$, can we also find the point on the curve where the slope is zero?

 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, $\left(x(2),y(2) \right)$ ... Thanks from rath

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