Calculus Calculus Math Forum

 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

Joined: Jul 2011
From: Texas

Posts: 3,094
Thanks: 1677

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 Tags differentiation, line, question, slope 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 Tooperoo Calculus 9 May 7th, 2012 01:17 AM ads98765 Calculus 7 February 15th, 2012 04:10 PM SH-Rock Calculus 12 September 29th, 2010 09:12 PM zoopie2 Calculus 1 August 26th, 2010 05:53 PM Myrtho Calculus 1 March 11th, 2009 03:37 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      