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The distance in metres travelled by a body in π‘ seconds is π equals nine π‘ squared plus five π‘ plus seven.
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What is the rate of change of π with respect to π‘ when π‘ is equal to 11?
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Another way to consider the rate of change of π is as the change in π over a given period of time.
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And the function we use to find the rate of change at a given time, π‘ one, is the limit as β tends to zero of π of π‘ one plus β minus π of π‘ one all over β.
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Letβs begin then by finding π of π‘ one plus β.
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Now, π is a function in π‘.
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So we can say π of π‘ is equal to nine π‘ squared plus five π‘ plus seven.
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So to find π of π‘ one plus β, we replace every π‘ with π‘ one plus β in this equation.
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And we get nine multiplied by π‘ one plus β squared plus five multiplied by π‘ one plus β plus seven.
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At this stage, it is sensible first to expand the brackets π‘ one plus β squared.
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We can write that as π‘ one plus β multiplied by π‘ one plus β.
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We then multiply the first term in each bracket.
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π‘ one multiplied by π‘ one is π‘ one squared.
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We multiply the outer terms.
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π‘ one multiplied by β is βπ‘ one.
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And when we multiply the inner terms, we get βπ‘ one again.
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Finally, we multiply the last term in each bracket.
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β multiplied by β is β squared.
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And then, we simplify the expression to get π‘ one squared plus two βπ‘ one plus β squared.
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All thatβs left to do here is to expand these two brackets.
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Weβre going to multiply everything in this first bracket by nine and everything in the second bracket by five.
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And when we do, we get nine π‘ one squared plus 18βπ‘ plus β squared plus five π‘ one plus five β plus seven.
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π of π‘ one is a little bit simpler.
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We replace π‘ with π‘ one.
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And we get nine π‘ one squared plus five π‘ one plus seven.
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And we now see the rate of change is found by the limit as β tends to zero of this rather nasty-looking expression.
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We put a pair of brackets around that second expression to remind us that we needed to subtract five π‘ squared as well as seven.
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Then, nine π‘ one squared minus nine π‘ one squared is zero.
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Five π‘ one minus five π‘ one is zero.
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And seven minus seven is zero.
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And it becomes 18βπ‘ one plus β squared plus five β all over β.
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And weβll divide through by this β.
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And we see that our expression simplifies to 18π‘ one plus β plus five.
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And we can now evaluate this as β tends to zero.
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As β tends to zero, our expression becomes 18π‘ one plus five.
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And all that remains is to substitute π‘ is equal to 11 into this expression.
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At π‘ is equal to 11, the rate of change is 18 multiplied by 11 plus five, which is 203.
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Now, no units are required here.
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But letβs think about what they might be.
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Weβre finding the rate of change of distance over time.
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Thatβs the same as speed.
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And since the distance travelled is in metres and the time is in seconds, the units, if we required them, would be metres per seconds.
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We donβt know.
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So we say the rate of change at π‘ is equal to 11 is 203.