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A particle is moving in a straight line such that its displacement π after π‘ seconds is given by π equals three π‘ cubed minus 54π‘ squared plus 38π‘ meters, where π‘ is greater than or equal to zero.
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Determine the time interval during which the velocity of the particle is increasing.
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This question asks us something about the velocity of the particle.
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We have the displacement of this particle π in terms of time π‘.
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So all we have to do to find the velocity of the particle is to differentiate this with respect to time.
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Okay, so letβs differentiate.
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Calling the velocity π£, π£ is the derivative with respect to time π‘ of displacement π .
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And π is three π‘ cubed minus 54π‘ squared plus 38π‘.
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So differentiating term by term using the fact that the derivative π by ππ‘ of π times π‘ to the π is π times π times π‘ to the π minus one, we get that the velocity is 9π‘ squared minus 108π‘ plus 38.
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Now that we have the velocity, we can start to consider the problem weβve been given to determine the time interval during which this velocity is increasing.
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One way to do this would be to sketch the graph of 9π‘ squared minus 108π‘ plus 38 again itβs a π‘.
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Hopefully, you know how to do this.
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But there is another way we can solve this.
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The velocity of the particle is increasing if the rate of change of velocity with respect to time β that is ππ£ by ππ‘ β is greater than zero.
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More generally, a general function π is increasing when the derivative π prime of π₯ is greater than zero.
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So letβs find ππ£ by ππ‘ by differentiating again.
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Again, we differentiate term by term using the power rule to get 18π‘ minus 108.
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So when is ππ£ by ππ‘ greater than zero?
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Well, ππ£ by ππ‘ is 18π‘ minus 108.
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So this is when 18π‘ minus 108 is greater than zero, which is when 18π‘ is greater than 108, which is when π‘ is greater than six.
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Writing this in interval notation, this is the open interval from six to infinity, including neither endpoint and so using a parenthesis rather than square brackets.
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This is the time interval during which the velocity of the particle whose displacement is given by π equals three π‘ cubed minus 54π‘ squared plus 38π‘ meters, where π‘ is greater than zero is increasing.
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This corresponds to where the derivative of velocity with respect to time is greater than zero.
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And of course, we know the derivative of a velocity with respect to time by another name.
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Itβs the acceleration of the particle.
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And so another way of phrasing this question would be to ask where the acceleration of the particle is greater than zero.