Author Topic: Inequality help (Read 1256 times) Tweet Share

dekoyl · « **on:** July 15, 2009, 04:13:28 pm »

A particle is moving in a straight line and its displacement from the origin at any time $t$ is given by $x=cos(2t)+4cos(t)$ . In the time interval $0<t<2\pi$ find when $\frac{d^2x}{dt^2}>0$ .

At first I had:
$2cos^2(t)-1+cos(t)<0$
$cos(t)[2cos(t)+1]<1$
But I couldn't extract the right answers from that. However, when I had:

$2cos^2(t)-1+cos(t)<0$
$\left(2cos(t)-1\right)\left(cos(t)+1\right)<0$
I could get the answers.

Is there an explanation for which method to use? I think I encountered something like this in methods whereby some solutions would be missing if I factorised a specific way.

The answer is $\frac{\pi}{3}<t<\frac{5\pi}{3}$ except $\pi$ . Thanks a lot.

humph · « **Reply #1 on:** July 15, 2009, 04:27:34 pm »

Quote from: dekoyl on July 15, 2009, 04:13:28 pm

A particle is moving in a straight line and its displacement from the origin at any time $t$ is given by $x=cos(2t)+4cos(t)$ . In the time interval $0<t<2\pi$ find when $\frac{d^2x}{dt^2}>0$ .

At first I had:
$2cos^2(t)-1+cos(t)<0$
$cos(t)[2cos(t)+1]<1$
But I couldn't extract the right answers from that. However, when I had:

$2cos^2(t)-1+cos(t)<0$
$\left(2cos(t)-1\right)\left(cos(t)+1\right)<0$
I could get the answers.

Is there an explanation for which method to use? I think I encountered something like this in methods whereby some solutions would be missing if I factorised a specific way.

The answer is $\frac{\pi}{3}<t<\frac{5\pi}{3}$ except $\pi$ . Thanks a lot.

$x(t) = \cos 2t + 4\cos t$
$\Longrightarrow x'(t) = -2\sin 2t - 4\sin t$
$\Longrightarrow x''(t) = -4\cos 2t - 4\cos t = -8 \cos^2 t + 4 - 4\cos t = -4(2\cos t - 1)(\cos t + 1)$
Thus $x''(t) > 0$ when
(1) $2\cos t - 1 > 0$ and $\cos t + 1 < 0$ ,
OR
(2) $2\cos t - 1 < 0$ and $\cos t + 1 > 0$ .
Note that $\cos t + 1 < 0$ is impossible, so we only need to consider case (2). In this case, we are left with the conditions $\cos t < 1/2$ and $\cos t > -1$ . The first condition implies that $\pi/3 < t < 5\pi/3$ , and the second condition implies that $t \neq \pi$ .

Best way to solve this type of question is to factorise and have an inequality of the form $a_1 \cdots a_n > 0$ , and then consider the possible combinations of $a_i > 0$ or $a_i < 0$ for which this inequality holds (so more of a combinatorial argument).

TrueTears · « **Reply #2 on:** July 15, 2009, 04:38:25 pm »

Well when I first saw the equation I thought of subbing u = cos(t) straight away because that way you can get a quadratic.

Although by looking at your first method I don't see how you can solve for t after you get $cos(t)[2cost(t)+1] < 1$ . What's the next step?

So I just did substitution.

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Author Topic: Inequality help (Read 1256 times) Tweet Share

dekoyl

Inequality help

humph

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TrueTears

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