Question

I'm quite confused with the racetrack principle and need help on this problem.

Suppose that f(t) is continuous and twice-differentiable for t>= 0. Further suppose f''(t) >= 3 for all t>= 0 and f(0) = f'(0) = 0.

Using the Racetrack Principle, what linear function g(t) can we prove is less than f'(t) (for t>= 0)?
g(t) =

Then, also using the Racetrack Principle, what quadratic function h(t) can we prove is less than than f(t) (for t>= 0)?
h(t) =

Answer 1

do you understand the racetrack principle?

Answer 2

hey my name's Peter toooo lol. But no not really. My teacher briefly went over it right before class ended.. But I looked it up online to try and better understand it but I don't really get it.

Answer 3

Race track principle says that if two functions are equal at \(t=0\), then the one which has a greater derivative will be greater.

Answer 4

In this case we're comparing \(f'(t)\) and \(g'(t)\). So we make sure that \(g(0)=f'(0)\) and that \(f''(t)\geq g'(t)\)

Answer 5

\[
g(t) = at+b
\]Since it is a line.\[
g'(t) = a
\]

Answer 6

\[
f''(t)\geq 3\geq g'(t)\implies 3\geq a
\]So let \(a=3\).\[
f'(0)=0=g(0)=3(0)+b\implies b=0
\]

Answer 7

So that means \[
g(t) = 3t
\]

Answer 8

Do something similar for \(h(t)\) starting with \[
h(t) = at^2+bt+c
\]

Answer 9

\[
h(0) = f(0) \implies c=0
\]

Answer 10

So \[
h(t) = at^2+bt
\]

Answer 11

Can you do it now?

Answer 12

So the derivative of h(t) is 2at+b. Then do I plug in 3 for a and 0 for t..?