# Parametric 2'nt deratives

## Homework Statement

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

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?

## Answers and Replies

Defennder
Homework Helper
M=2t/(2t+1)=0
I assume that by M you mean $$\frac{d^2y}{dx^2}$$. 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.