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The power of a machine is given by the relation π is equal to three π‘ plus seven, where π‘ is the time elapsed in seconds.
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Find the work done by the machine in the first eight seconds.
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In this question, weβre given an expression for the power of a machine π in terms of time π‘.
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It is equal to three π‘ plus seven.
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We are asked to find the work done by the machine.
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Since the power supplied by a force is the time derivative of the work done, we can find an expression for the work done by integrating our expression for the power with respect to π‘.
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In this question, π€ is equal to the integral of three π‘ plus seven with respect to π‘.
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As we are interested in the work done by the machine in the first eight seconds, we will have a definite integral, where our lower limit is equal to zero and our upper limit is eight.
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Integrating term by term gives us three π‘ squared over two plus seven π‘.
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And as weβre dealing with a definite integral, there will be no constant of integration.
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Next, we can substitute in the upper and lower limits and then find a difference between these values.
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This is equal to 96 plus 56 minus zero plus zero, which in turn is equal to 152.
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Since we are not given the units of power in the question, we can conclude that the work done by the machine in the first eight seconds is 152 work units.
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An alternative method to answer this question would be to sketch the relation π is equal to three π‘ plus seven.
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Our function for power is linear, so we have a straight-line graph with π¦-intercept equal to seven and slope or gradient equal to three.
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The work done is equal to the area enclosed by our line, the π₯-axis, and the vertical lines π‘ equals zero and π‘ equals eight.
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This forms a trapezoid or trapezium.
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And we know the area of this is equal to π plus π over two multiplied by β, where π and π are the lengths of the parallel sides and β is the perpendicular distance between them.
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When π‘ is equal to zero, π is equal to three multiplied by zero plus seven, which is equal to seven.
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This is the π¦-intercept as already mentioned.
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When π‘ is equal to eight, π is equal to three multiplied by eight plus seven, which is equal to 31.
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The area of our trapezoid is therefore equal to seven plus 31 divided by two multiplied by eight.
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This is equal to 19 multiplied by eight, which is equal to 152.
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And this confirms that the work done by the machine in the first eight seconds is 152 work units.