The distance in metres travelled by a body in 𝑡 seconds is 𝑆 equals nine 𝑡 squared plus five 𝑡 plus seven. What is the rate of change of 𝑆 with respect to 𝑡 when 𝑡 is equal to 11?
Another way to consider the rate of change of 𝑆 is as the change in 𝑆 over a given period of time. And the function we use to find the rate of change at a given time, 𝑡 one, is the limit as ℎ tends to zero of 𝑓 of 𝑡 one plus ℎ minus 𝑓 of 𝑡 one all over ℎ.
Let’s begin then by finding 𝑓 of 𝑡 one plus ℎ. Now, 𝑆 is a function in 𝑡. So we can say 𝑓 of 𝑡 is equal to nine 𝑡 squared plus five 𝑡 plus seven. So to find 𝑓 of 𝑡 one plus ℎ, we replace every 𝑡 with 𝑡 one plus ℎ in this equation. And we get nine multiplied by 𝑡 one plus ℎ squared plus five multiplied by 𝑡 one plus ℎ plus seven. At this stage, it is sensible first to expand the brackets 𝑡 one plus ℎ squared.
We can write that as 𝑡 one plus ℎ multiplied by 𝑡 one plus ℎ. We then multiply the first term in each bracket. 𝑡 one multiplied by 𝑡 one is 𝑡 one squared. We multiply the outer terms. 𝑡 one multiplied by ℎ is ℎ𝑡 one. And when we multiply the inner terms, we get ℎ𝑡 one again. Finally, we multiply the last term in each bracket. ℎ multiplied by ℎ is ℎ squared. And then, we simplify the expression to get 𝑡 one squared plus two ℎ𝑡 one plus ℎ squared.
All that’s left to do here is to expand these two brackets. We’re going to multiply everything in this first bracket by nine and everything in the second bracket by five. And when we do, we get nine 𝑡 one squared plus 18ℎ𝑡 plus ℎ squared plus five 𝑡 one plus five ℎ plus seven. 𝑓 of 𝑡 one is a little bit simpler. We replace 𝑡 with 𝑡 one. And we get nine 𝑡 one squared plus five 𝑡 one plus seven.
And we now see the rate of change is found by the limit as ℎ tends to zero of this rather nasty-looking expression. We put a pair of brackets around that second expression to remind us that we needed to subtract five 𝑡 squared as well as seven. Then, nine 𝑡 one squared minus nine 𝑡 one squared is zero. Five 𝑡 one minus five 𝑡 one is zero. And seven minus seven is zero. And it becomes 18ℎ𝑡 one plus ℎ squared plus five ℎ all over ℎ. And we’ll divide through by this ℎ. And we see that our expression simplifies to 18𝑡 one plus ℎ plus five.
And we can now evaluate this as ℎ tends to zero. As ℎ tends to zero, our expression becomes 18𝑡 one plus five. And all that remains is to substitute 𝑡 is equal to 11 into this expression. At 𝑡 is equal to 11, the rate of change is 18 multiplied by 11 plus five, which is 203. Now, no units are required here. But let’s think about what they might be. We’re finding the rate of change of distance over time. That’s the same as speed. 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. We don’t know.
So we say the rate of change at 𝑡 is equal to 11 is 203.