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Ad libitum [116K]
3 years ago
7

I'm quite confused with the racetrack principle and need help on this problem. . . Suppose that f(t) is continuous and twice-dif

ferentiable 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) = .
Mathematics
1 answer:
Hunter-Best [27]3 years ago
6 0

Race track principle says that if two functions are equal at <span>t=0</span>, then the one which has a greater derivative will be greater.

In this case we're comparing <span><span>f′</span>(t)</span> and <span><span>g′</span>(t)</span>. So we make sure that <span>g(0)=<span>f′</span>(0)</span> and that <span><span>f′′</span>(t)≥<span>g′</span>(t)

</span>


<span>g(t)=at+b</span>


Since it is a line.
<span><span>g′</span>(t)=a</span>


<span><span>f′′</span>(t)≥3≥<span>g′</span>(t)⟹3≥a</span>

So let <span>a=3</span>.

<span><span>f′</span>(0)=0=g(0)=3(0)+b⟹b=0


</span>So that means
<span>g(t)=3t

</span>Do something similar for <span>h(t)</span><span> starting with
</span><span>h(t)=a<span>t2</span>+bt+c

</span><span>h(0)=f(0)⟹c=0

</span>

So

<span>h</span><span>(</span><span>t</span><span>)</span><span>=</span><span>a</span><span>t2</span><span>+</span><span>b</span><span>t</span>

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