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May 14th, 2016, 06:34 AM   #1
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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
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May 14th, 2016, 07:19 AM   #2
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$x=t^3 \implies \dfrac{dx}{dt}=3t^2$

$y=t^2-4t \implies \dfrac{dy}{dt}=2t-4$

$\dfrac{dy}{dx}=\dfrac{2t-4}{3t^2}$

a tangent line parallel to the x-axis would have a slope equal to zero, correct?

What value of $t$ would make $\dfrac{dy}{dx}=0$ ?
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May 14th, 2016, 07:26 AM   #3
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t = 2 ?
$\displaystyle 2(2)-4/3(2^3) $
$\displaystyle 0/24 = 0 $
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May 14th, 2016, 07:59 AM   #4
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Quote:
Originally Posted by rath View Post
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?
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May 14th, 2016, 08:06 AM   #5
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(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.
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May 14th, 2016, 08:25 AM   #6
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the point is on the curve, $\left(x(2),y(2) \right)$ ...
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