Question Video: Finding the Rate of Change of a Polynomial Function Representing the Distance Travelled by a Body in a Certain Time Mathematics

Video Transcript

The distance in meters traveled by a body in 𝑡 seconds is 𝑠 is equal to nine 𝑡 squared plus five 𝑡 plus seven. What is the instantaneous 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 that we use to find the rate of change at a given time 𝑡 sub one is the limit as ℎ tends to zero of 𝑓 of 𝑡 sub one plus ℎ minus 𝑓 of 𝑡 sub one all divided by ℎ.

Let’s begin by finding 𝑓 of 𝑡 sub one plus ℎ. Since 𝑠 is a function of 𝑡, we can rewrite this as 𝑓 of 𝑡 is equal to nine 𝑡 squared plus five 𝑡 plus seven. So, to find 𝑓 of 𝑡 sub one plus ℎ, we replace every 𝑡 with 𝑡 sub one plus ℎ in this equation. We have nine multiplied by 𝑡 sub one plus ℎ squared plus five multiplied by 𝑡 sub one plus ℎ plus seven.

Noting that 𝑡 sub one plus ℎ squared is equal to 𝑡 sub one squared plus two 𝑡 sub one ℎ plus ℎ squared, we can distribute the parentheses as follows. Our expression becomes nine 𝑡 sub one squared plus 18𝑡 sub one ℎ plus nine ℎ squared plus five 𝑡 sub one plus five ℎ plus seven. We can now find an expression for 𝑓 of 𝑡 sub one. Replacing 𝑡 with 𝑡 sub one in our function, we have nine 𝑡 sub one squared plus five 𝑡 sub one plus seven.

We can now find the rate of change by firstly subtracting these two expressions. When we do this, the terms nine 𝑡 sub one squared, five 𝑡 sub one, and the constant seven cancel. The rate of change is therefore equal to the limit as ℎ tends to zero of 18𝑡 sub one ℎ plus nine ℎ squared plus five ℎ divided by ℎ. As ℎ tends to zero and is therefore never equal to zero, we can divide through by ℎ.

We have the limit as ℎ tends to zero of 18𝑡 sub one plus nine ℎ plus five. Next, we use direct substitution of ℎ equals zero. This gives us 18𝑡 sub one plus five. We now have an expression for the rate of change. However, we’re asked for the instantaneous rate of change when 𝑡 is equal to 11. To calculate this, we multiply 18 by 11 and then add five. This is equal to 203.

The instantaneous rate of change of 𝑠 with respect to 𝑡 when 𝑡 is equal to 11 is 203. Since the distance 𝑠 was measured in meters and the time 𝑡 in seconds, we could add units for the rate of change of meters per second.