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# Graphs and Differentiation

Graphs and Differentiation
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Nie wieder prokastinieren mit unseren Lernerinnerungen. The gradient of a graph is essentially the steepness of a graph at a particular point. If we sketch a function we are able to visualise this.

Let's take a look at a simple linear equation to work out the gradient:

Let$y=3x+4$, the gradient is 3 as the gradient of a graph of$y=mx+c$ is$m$.

What happens when we have a curve and not a straight line graph? Well, instead of just one point which has a real value for a gradient we now have a function of our gradient. Let's look at an example of a curve to explain this:

Let$y={x}^{2}$. Find the value of the gradient at the point$\left(2,4\right)$. ## What are minimum and maximum points?

Now, when a graph has a positive value for$\frac{dy}{dx}$, it is a graph of positive gradient and when a graph has a negative value for$\frac{dy}{dx}$, it has a negative gradient.

The blue line signifies a negative gradient, the green line a zero gradient and the purple line signifies a positive gradient.

However, there are times when the value of$\frac{dy}{dx}$is neither positive nor negative and is zero.

This is when the graph has a maximum or minimum point.

As an example below we can see$y={x}^{2}$ has a minimum point at$\left(0,0\right)$ and$y=-{x}^{2}$ has a maximum point at$\left(0,0\right)$. Two graphs with the same turning point

$\left(0,0\right)$ is the minimum point of$y={x}^{2}$and the maximum point of$y=-{x}^{2}$.

To find whether it is maximum or minimum, then calculate the second derivative$\frac{{d}^{2}y}{d{x}^{2}}$.

If the value of$\frac{{d}^{2}y}{d{x}^{2}}<0\phantom{\rule{0ex}{0ex}}$ , then it is a maximum point.

If the value of$\frac{{d}^{2}y}{d{x}^{2}}>0$, then it is a minimum point.

Otherwise, it is a point of inflection. This means it changes direction. Below is an example of a point of inflection with the graph$y={x}^{3}$at$\left(0,0\right)$. Graph with point of inflection

The graph$y={x}^{3}$containing a stationary point of inflection.

Let's look at an example of this.

Find the minimum point of$y=3{x}^{2}-6x+1$. Let's look at an example where we have to find out whether it is a minimum or maximum.

Find the x-values of the stationary points of$y=13{x}^{3}+32{x}^{2}+2x+5$. Determine the nature of these stationary points.

SOLUTION:

$\frac{dy}{dx}=39{x}^{2}+64x+2$

By using the quadratic formula we can find our x-values:

$\frac{-64+\sqrt{{64}^{2}-\left(4×39×2\right)}}{78}=-0.03186889747\phantom{\rule{0ex}{0ex}}\frac{-64-\sqrt{{64}^{2}-\left(4×39×2\right)}}{78}=-1.60915674355$

To determine the nature of our stationary points we have to compute the second derivative.

$\frac{{d}^{2}y}{d{x}^{2}}=78x+64$

Substitute in both these x-values and comment on whether or not they are negative or positive:

$78\left(-1.60915674355\right)+64=140.390843>0\phantom{\rule{0ex}{0ex}}78\left(-0.03186889747\right)+64=141.968131>0\phantom{\rule{0ex}{0ex}}$

These are both positive therefore they are minimum points.

## What is a geometric interpretation of this example?

When$y={x}^{2}$, the gradient at the point$\left(2,4\right)$ is$4$.

The key thing from the example above is that the gradient is $4$.

This is the gradient of the curve at a specific point. It is therefore the equation of the tangent line to the curve at the point $\left(2,4\right)$.

A tangent is a line which touches a curve at just one point, not intersecting it. Tangent lines to the graph y=x2

The blue lines are all tangent to the purple curve in the middle.

This straight line graph touches this line. We will now look at how to find the equation of the tangent.

## Equation of a tangent

All tangents are straight line graphs in the form of:$y=mx+c$ meaning that we need to calculate the values of $m$ and $c$.

The value of $m$is the value of the gradient so we take that from finding $\frac{dy}{dx}$ of the function.

The value of $c$ is found by substituting in the $x$ and $y$ values with our newly found value of $m$.

Let's look at an example.

Find the equation of the tangent to the graph $y={x}^{3}+4{x}^{2}$ at the point $\left(4,128\right)$.

SOLUTION:

Firstly let's calculate the value of the gradient $m$, we can use $\frac{dy}{dx}$ to do this.

$\frac{dy}{dx}=3{x}^{2}+8x$Now substitute the value $x=4$:$\frac{dy}{dx}=3\left({4}^{2}\right)+8\left(4\right)=48+32=80$Or so $y=80x+c$. Now to find c, we need to substitute $x=4$ and $y=128$. So:$128=80\left(4\right)+c\phantom{\rule{0ex}{0ex}}128=320+c\phantom{\rule{0ex}{0ex}}c=128-320=-192$So our equation is$y=80x-192$

Now let's look at a more complex example involving trigonometric functions.

Find the equation of the tangent to the graph $y=\mathrm{tan}\left(x\right)$ at the point $\left(\frac{\pi }{4},1\right)$.

SOLUTION:

Firstly let's calculate the value of the gradient $m$, we can use $\frac{dy}{dx}$ to do this.

$\frac{dy}{dx}=se{c}^{2}\left(x\right)$Now substitute the value $x=\frac{\pi }{4}$:$\frac{dy}{dx}=se{c}^{2}\left(\frac{\pi }{4}\right)=2$

So $y=2x+c$

Now to find $c$, we need to substitute $x=\frac{\pi }{4}$ and $y=1$. So $1=2\left(\frac{\pi }{4}\right)+c\phantom{\rule{0ex}{0ex}}1=\frac{\pi }{2}+c\phantom{\rule{0ex}{0ex}}c=1-\frac{\pi }{2}$So our equation is $y=2x+\left(1-\frac{\pi }{2}\right)$

## Equation of a normal

The normal is the line perpendicular to the tangent. Below we can see a curve, an equation of a tangent line and an equation of a normal line. Lines intersecting with normal, curve, tangent.

The two red lines represent the curve and the tangent and the blue line is the equation of the normal.

So to find the equation of the normal, firstly find the gradient of the tangent. Then use the fact that the gradients of two perpendicular lines equal -1. Then find the equation of the normal.

Let's look at the examples above but with normals to make sense of it.

Find an equation of the normal line of the graph$y={x}^{3}+4{x}^{2}$ at the point$\left(4,128\right)$.

SOLUTION:

Firstly let's calculate the value of the gradient $m$, we can use $\frac{dy}{dx}$ to do this.

$\frac{dy}{dx}=3{x}^{2}+8x$

Now substitute the value $x=4$:

$\frac{dy}{dx}=3\left({4}^{2}\right)+8\left(4\right)=48+32=80$So$y=80x+c$. Now let's find the gradient of the normal. We know that when we multiply the gradient of the normal and $80$ we get $-1$. So the gradient of the normal is $-\frac{1}{80}$. Therefore, the equation of the normal is$y=-\frac{1}{80}x+c$ Now we substitute $y=128andx=4$ we get:$128=-\frac{1}{80}\left(4\right)+c\phantom{\rule{0ex}{0ex}}c=128.05\phantom{\rule{0ex}{0ex}}y=-\frac{1}{80}x+128.05$

Find the equation of the normal to the graph $y=\mathrm{tan}\left(x\right)$ at the point $\left(\frac{\pi }{4},1\right)$.

SOLUTION: Firstly let's calculate the value of the gradient m, we can use $\frac{dy}{dx}$to do this.$\frac{dy}{dx}=se{c}^{2}\left(x\right)$Now substitute the value $x=\frac{\pi }{4}$:$\frac{dy}{dx}=se{c}^{2}\left(\frac{\pi }{4}\right)=2$So $y=2x+c$Now let's find the gradient of the normal. We know that when we multiply the gradient of the normal and 2 we get -1. So the gradient of the normal is $-\frac{1}{2}$. Therefore, the equation of the normal is$y=-\frac{1}{2}x+c$Now substitute $y=1$ and $x=\frac{\pi }{4}$. We get$1=-\frac{1}{2}\left(\frac{\pi }{4}\right)+c$$c=1+\frac{\pi }{8}$So our equation is $y=-\frac{1}{2}x+\left(1+\frac{\pi }{8}\right)$

As we can see with both trigonometric examples we don't simplify numbers involving$\pi$. This is because we want everything in exact form, and$\pi$ is an irrational number.

## Graphs and Differentiation - Key takeaways

• Graphs can be used to depict gradients.

• Gradients are calculated through differentiation and are the m value in linear equations y = mx + c.

• Equations of tangents are used to show the equation of the line with the same gradient as the graph.

• Equation of a normal is the equation of the line perpendicular to the tangent line.

T

When you differentiate you calculate the gradient of the tangent at that specific point to the graph.

The first derivative shows the gradient at that specific point. The second derivative shows us what type of stationary point we have.

If the gradient of the tangent line is positive or negative. 60%

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