Lesson

We've learnt so much already about tangents and gradients of curves. You must be wondering whether there's an easier way to find gradients rather than drawing tangent lines and measuring their slope. Of course there is!

But before we get there we want to play the part of Mathematicians like Liebniz and Newton and discover the connection for ourselves. By doing this, when we begin formalising concepts like gradients of curves and gradient functions, we'll have a solid foundation with which to understand it all.

To illustrate this process, we'll investigate the gradient of the curve $f(x)=x^2$`f`(`x`)=`x`2 at the point $x=1$`x`=1.

Before we begin, you might like to go back and read through this investigation so you get a proper feel for what we're doing.

If you've read through it, you'll see that what we're doing is finding the gradient of chords, drawing smaller and smaller chords until we get closer and closer to $x=1$`x`=1.

Why not just draw a tangent you might ask?

Well, we could in this case as the gradient at $x=1$`x`=1 is going to turn out to be an integer, but let's say we wanted to know the gradient at $x=2.7$`x`=2.7. The tangent here won't be as easy to draw, nor will it be easy to calculate. So we'll use a method that works to approximate the gradient at ANY point on ANY curve.

We'll begin by calculating the gradient of the chord drawn between $x=1$`x`=1 and $x=2$`x`=2. We'll follow this with smaller and smaller chords, and tabulate our results.

As we can see, as $h$`h` gets smaller and closer to $0$0, the gradient of the curve at $x=1$`x`=1 approaches a value of $2$2.

To take this further, conduct this limiting chord process for a few different $x$`x` values and complete a table similar to the one given below.

Can you see a pattern in the values in the second row? You're getting close now to uncovering the important ideas of calculus.

Consider the function $f\left(x\right)=x^3$`f`(`x`)=`x`3

By filling in the table of values, complete the limiting chord process for $f\left(x\right)=x^3$

`f`(`x`)=`x`3 at the point $x=1$`x`=1.Answer up to 4 decimal places if required.

$a$ `a`$b$ `b`$h=b-a$ `h`=`b`−`a`$\frac{f\left(b\right)-f\left(a\right)}{b-a}$

`f`(`b`)−`f`(`a`)`b`−`a`$1$1 $2$2 $1$1 $\editable{}$ $1$1 $1.5$1.5 $\editable{}$ $\editable{}$ $1$1 $1.1$1.1 $\editable{}$ $\editable{}$ $1$1 $1.05$1.05 $\editable{}$ $\editable{}$ $1$1 $1.01$1.01 $\editable{}$ $\editable{}$ $1$1 $1.001$1.001 $\editable{}$ $\editable{}$ $1$1 $1.0001$1.0001 $\editable{}$ $\editable{}$ The instantaneous rate of change of $f\left(x\right)$

`f`(`x`) at $x=1$`x`=1 is

Consider the function $f\left(x\right)=2x^2$`f`(`x`)=2`x`2

By filling in the table of values, complete the limiting chord process for $f\left(x\right)=2x^2$

`f`(`x`)=2`x`2 at the point $x=3$`x`=3.$a$ `a`$b$ `b`$h=b-a$ `h`=`b`−`a`$\frac{f\left(b\right)-f\left(a\right)}{b-a}$

`f`(`b`)−`f`(`a`)`b`−`a`$3$3 $4$4 $1$1 $\editable{}$ $3$3 $3.5$3.5 $\editable{}$ $\editable{}$ $3$3 $3.1$3.1 $\editable{}$ $\editable{}$ $3$3 $3.05$3.05 $\editable{}$ $\editable{}$ $3$3 $3.01$3.01 $\editable{}$ $\editable{}$ $3$3 $3.001$3.001 $\editable{}$ $\editable{}$ $3$3 $3.0001$3.0001 $\editable{}$ $\editable{}$ Provide all answers up to $4$4 decimal places if required.

The limiting chord process has been used to calculate the instantaneous rate of change for each value of $x$

`x`given in the table. Complete the missing values.$x$ `x`$1$1 $2$2 $3$3 $4$4 $5$5 Instantaneous rate of change of $f\left(x\right)$

`f`(`x`) at $x$`x`.$4$4 $8$8 $\editable{}$ $16$16 $\editable{}$ From the results of the previous parts, we can deduce that the instantaneous rate of change of $f\left(x\right)$

`f`(`x`) for any given value of $x$`x`is:

Choose and apply a variety of differentiation, integration, and antidifferentiation techniques to functions and relations, using both analytical and numerical methods

Apply differentiation methods in solving problems