Parametric 2'nt deratives

  • Thread starter thharrimw
  • Start date
  • #1
115
0

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?
 

Answers and Replies

  • #2
Defennder
Homework Helper
2,591
5
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.
 

Related Threads on Parametric 2'nt deratives

Replies
5
Views
936
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
2K
Replies
2
Views
2K
Replies
1
Views
56K
Replies
1
Views
1K
Replies
6
Views
4K
Replies
4
Views
1K
Replies
2
Views
760
Replies
1
Views
2K
Top