Question

Find the values of t when the function defined by x=4t^3+1 and y=3t^2-7t is concave up.

Answer 1

\[x=4t^3+1 \]
\[\frac{ dx }{ dt }=12t^2\]

and then
\[y=3t^2-7t\]
\[\frac{ dy }{ dt }=6t-7\]

now to obtain dy/dx
\[\frac{ dy }{ dx }=\frac{ 6t-7 }{ 12t^2 }\]

Answer 2

now how do i go about looking for the values where it concave up? I have to take the second derivative but what is the x value we are after?

Answer 3

you need to find values of t such that
d^2y/dx^2 > 0

Answer 4

but i cant do d^2y/dx^2 because my equation is written as t as a variable

Answer 5

you can use differential notation to derive a formula

Answer 6

can u plz show me

Answer 7

\[d^2y/dx^2=\frac{ -72t^2+168t }{ (12t^2)^2 }\]

Answer 8

$$ \large{\frac{d^2y}{dx^2 } =\frac{d}{dx}\left( \frac{dy/dt}{dx/dt} \right)
= \frac{d}{dt} \left( \frac{dy/dt}{dx/dt} \right) \frac{dt}{dx}
\\ = \frac{\Large \frac{d}{dt} \left( \frac{dy/dt}{dx/dt} \right) }{dx/dt}
}$$ verify that all the differentials cancel and work out

Answer 9

so then essentially what i need do it divide the above second differential by 12t^2 because that is dx/dt

Answer 10

right. take the derivative of (6t - 7) / 12t^2 , and then divide by 12t^2

Answer 11

do you think i need to go and make it greater than zero as well?

Answer 12

we need the final dy^2 / dx^2 in terms of t, then find t where its greater than zero

Answer 13

we could always explicitly solve for y in terms of x
it might be easier
x=4t^3 + 1
y = 3t^2 -7t

solve for t in first equation
t = ((x - 1)/4 ) ^(1/3)
substitute into second equation
y = 3 ( ((x - 1)/4 ) ^(1/3) ] ^2 - 7( (x - 1)/4 ) ^(1/3)
y = 3( (x-1)/4 )^(2/3) - 7(( x-1)/4)^(1/3)

we can see they are equivalent here https://www.desmos.com/calculator/6wh7bsuwt8

Answer 14

It looks like using t is easier.
I get
dy^2 / dx^2 = (7 - 3t) / ( 75 t^5 )
solve this greater than zero
Test the real number line , using t = 0 and t = 7/3 as your 'critical points'

------ 0 +++++ 7/3 -------

Answer 15

it is concave up between t = 0 and t = 7/3
the hard part is getting the expression for the second derivative

Answer 16

cool i got that same answer too

Answer 17

thanks heaps

Answer 18

Your welcome :)