Differentiation

New Zealand

Level 7 - NCEA Level 2

Lesson

As we move past learning the skills of calculus and differentiation, we're ready to use our skills to solve applied problems.

The basics of these application-style problems involve using and manipulating the gradient function in each of the following ways:

- Determine the gradient at a given point on the curve
- Determine the equation of the tangent at a given point on the curve
- Determine the point(s) on a curve where a given gradient exists
- Using the information about gradients and the original function to determine unknown coefficients in the polynomial $f\left(x\right)$
`f`(`x`).

We've explored this idea earlier and we can review it with this worked example.

Find the gradient of $f\left(x\right)=16x^{-3}$`f`(`x`)=16`x`−3 at $x=2$`x`=2.

Denote this gradient by $f'\left(2\right)$`f`′(2).

We've also explored this idea earlier and it's worth taking a look at another example.

Determine the equation of the tangent to the curve $f\left(x\right)=x^3-2x$`f`(`x`)=`x`3−2`x` at the point where $x=-1$`x`=−1.

Since a tangent is a straight line, we know the equation of our tangent will be of the form $y=mx+c$`y`=`m``x`+`c`. Let's begin by finding m, the gradient. To do this we'll need the gradient function, so first we need to differentiate $f\left(x\right)$`f`(`x`).

$f'\left(x\right)=3x^2-2$`f`′(`x`)=3`x`2−2

Now let's use the gradient function to find the value of the gradient at $x=-1$`x`=−1.

$f'\left(-1\right)=3\left(-1\right)^2-2=1$`f`′(−1)=3(−1)2−2=1

We now know that $m=1$`m`=1and thus we have $y=x+c$`y`=`x`+`c`.

To find the value of $c$`c`, we need to substitute a point on the tangent into the equation. However, we only know one point, and that is where the tangent touches the curve. And so far all we know is the $x$`x`value. So let's find the $y$`y`value of the coordinate by using the original function.

$f\left(-1\right)=\left(-1\right)^3-2\left(-1\right)=1$`f`(−1)=(−1)3−2(−1)=1

So we have the point $\left(-1,1\right)$(−1,1).

We can now substitute this point to find the value of $c$`c`.

$1=1\times-1+c$1=1×−1+`c`

$c=2$`c`=2

Therefore the equation of the tangent is $y=x+2$`y`=`x`+2.

Let's review this with the following worked example.

Consider the function $f\left(x\right)=x^2+5x$`f`(`x`)=`x`2+5`x`.

Find the $x$

`x`-coordinate of the point at which $f\left(x\right)$`f`(`x`) has a gradient of $13$13.Hence state the coordinates of the point on the curve where the gradient is $13$13.

The function $f\left(x\right)=x^3+ax^2+bx+c$`f`(`x`)=`x`3+`a``x`2+`b``x`+`c` has a $y$`y`-intercept of $3$3. The function has stationary points at $x=1$`x`=1 and a root at $x=-3$`x`=−3. Determine the values of $a$`a`, $b$`b`and $c$`c`.

We can begin by using the information about the y-intercept.

$3=0^3+a\times0^2+b\times0+c$3=03+`a`×02+`b`×0+`c`

$c=3$`c`=3

To use the information about the stationary point, we will need to first find ourselves the gradient function.

$f'\left(x\right)=3x^2+2ax+b$`f`′(`x`)=3`x`2+2`a``x`+`b`

Substituting in the $x$`x`value of the stationary point, we obtain the equation:

$3+2a+b=0$3+2`a`+`b`=0 - equation 1

To use the information about the x-intercept, we can substitute this into the original function to obtain the equation:

$-27+9a-3b+3=0$−27+9`a`−3`b`+3=0 - equation 2

Solving simultaneously (either with the elimination method, substitution method or with a CAS calculator) we find that $a=1$`a`=1 and $b=-5$`b`=−5.