Differentiation

NZ Level 7 (NZC) Level 2 (NCEA)

Simple applications (using power rule only)

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.

Apply differentiation and anti-differentiation techniques to polynomials

Apply calculus methods in solving problems