Question

Instant fan/medal.
Can someone check this for me, I'm really struggling.

Determine the concavity of the curve with parametric equations x = 4 + t^2, y = t^2 + t^3 at t = 2.

dy/dx=(dy/dt)/(dx/dt)=(3t^2+2t) / (2x)

[d/dt ((3t^2+2t) / (2x))]/(2x)=(3x+1)/x

3*2+1 /2 = 7/2 >0 , so concave up.

Answer 1

Like you, I found it's convacave up but with a different result.
\[\frac{ dy }{ dx }=\frac{ \frac{ dy }{ dt } }{ \frac{ dx }{ dt } }=\frac{ 2t+3t ^{2} }{ 2t }=1+\frac{ 3 }{ 2 }t\]\[\frac{ d ^{2}y }{ dx ^{2} }=\frac{ d }{ dx }(\frac{ dy }{ dx })=\frac{ \frac{ d }{ dt }(\frac{ dy }{ dx }) }{ \frac{ dx }{ dt } }=\frac{ \frac{ 3 }{ 2 } }{ 2t }=\frac{ 3 }{ 4t }\]At t=2, \(\huge\frac{d^2y}{dx^2}=\frac{3}{8}\)