Slope and Concavity of a Parametric Function

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SUMMARY

The discussion focuses on finding the slope and concavity of the parametric function defined by x=t² and y=t²+t+1 at the point (0,0). The correct approach involves calculating the first derivative, M=2t/(2t+1), and then determining the second derivative using the chain rule. Participants clarify that to find the concavity, one must evaluate the second derivative at the appropriate t-value, which is derived from the equation y=0. The correct t-value for this case is t=0, leading to a concavity assessment based on the sign of the second derivative.

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Students studying calculus, particularly those focusing on parametric functions, as well as educators seeking to clarify concepts of slope and concavity in their teaching materials.

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Homework Statement


find the slope and concavity of the funtion at the given point.
x=t^2
Y=t^2+t+1
(0,0)

Homework Equations



t=x^(1/2)

The Attempt at a Solution


t=0 when x=0
x'=2t
y'=2t+1
M=2t/(2t+1)=0
for the second deritive would you take the deritive of 2t/(2t+1) devided by the deritive of 2t+1
once i find the second deritive i would plug in o for t and if it was + than it would be up -would be down. right?
 
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thharrimw said:
M=2t/(2t+1)=0
I assume that by M you mean [tex]\frac{d^2y}{dx^2}[/tex]. It should be y'/x' instead here.
for the second deritive would you take the deritive of 2t/(2t+1) devided by the deritive of 2t+1
Why should it be the derivative of y'=2t+1? How would you use the chain rule to determine a correct expression for d^2y/dx^2 ?
once i find the second deritive i would plug in o for t and if it was + than it would be up -would be down. right?
Not quite. You'll want to find the value of t for which x=y=0. For x=0,t=0 so that's correct. For y, setting t=0 gives y=1. Setting y=0 and solving for t gives you the correct t-values.
 

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