Rates of Change

NZ Level 7 (NZC) Level 2 (NCEA)

Average Vs Instantaneous Rates of Change

Lesson

Having just looked at average rate of change, it's time now to look at another type of rate of change, called instantaneous rate of change.

Instantaneous rate of change is the rate of change at that particular point, (not the average over a range of points).

At this stage you won't need to calculate the instantaneous rate of change (although that comes next), but for now, let's really get our head around on contextually, what the difference is.

We can have average speed, which is the average speed taken over a journey. For example if I drove from Canberra to Sydney in $5$5hrs, and the distance traveled is $340$340km, then my average speed is $\frac{340}{5}$3405 km/h = $68$68km/h. Now it's unlikely (and nearly impossible) to imagine that from the moment I started my car, to the moment I arrived in Sydney that I traveled at $68$68km/h the whole time. It's more likely that at some stages I was travelling $100$100km/h and others $40$40km/h through roadworks or any speed in between. It is these specific occasions, like the $40$40km/h past the road work sign, or the $100$100km/h through the police speed checkpoint that are the instantaneous rate of change. So we have the average rate of change, being a measure of the average speed for the trip, and the instantaneous rate of change being a measure of the speed at a point in the journey.

Consider the function $f\left(x\right)=2x+3$`f`(`x`)=2`x`+3 drawn below.

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Calculate the average rate of change between $x=-2$

`x`=−2 and $x=2$`x`=2.State the gradient of the tangent at the point $x=-2$

`x`=−2.

Consider the function $f\left(x\right)=x^3-3x^2+2x-1$`f`(`x`)=`x`3−3`x`2+2`x`−1

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Calculate the average rate of change between $x=2$

`x`=2 and $x=3$`x`=3.A tangent line has been drawn at $x=2$

`x`=2. Use the tangent to calculate the instantaneous rate of change at $x=2$`x`=2.

Consider the function $f\left(x\right)=2\left(3\right)^{-x}$`f`(`x`)=2(3)−`x`

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Calculate the average rate of change between $x=-1$

`x`=−1 and $x=0$`x`=0The tangent line at $x=-1$

`x`=−1 has been graphed, and its equation can be approximated by $y=-6.6x+6$`y`=−6.6`x`+6Use the tangent to calculate the instantaneous rate of change at $x=-1$

`x`=−1.

Apply differentiation and anti-differentiation techniques to polynomials

Apply calculus methods in solving problems