# Calculus of Curves

The image above shows a general curve. What could you say about its gradient? Does it change along the curve?

Using what we know about he gradient of a straight line, we can see where the gradient of a curve is positive, negative or zero by drawing tangent to the curve in difference places around the curve.

Notice that when the curve increases it has a positive gradient, when it decreases it has a negative gradient and when it turns around the gradient is zero. Another example is shown below here.

### Example

Sketch the gradient function of the curve below:

## Differentiation from First Principles

A function is differentiable if the gradient of the tangent can be found. Most functions are continuous, which means that they have a smooth unbroken line or curve. However, some have a gap, or discontinuity, in the graph (e.g. hyperbola). This can be shown by a asymptote or a 'hole' in the graph. We cannot find the gradient of a tangent to the curve at a point that doesn't exist! So the function is not differentiable at the point of discontinuity.

A function y=f(x) is differentiable at the point x=a if the derivative exists at that point. This can only happen if the function is continuous and smooth at x=a.

### Examples

1. Find all points where the function below is not differentiable.

The function is not differentiable at points A and B since there are sharp corners and the curve is not smooth at these points. It is also not differentiable at point C since the function is discontinuous at this point.

2. Is the function $f(x)=\begin{cases} { x }^{ 2 }\qquad & for\quad x\ge 1 \\ 3x-2\qquad & for\quad x<1 \end{cases}$ differentiable at all points?

## Limits

To differentiate from first principles, a limit is used when we want to move as close as we can to something. Often this is to find out where a function is near a gap or discontinuous point. In this topic, it is used when we want to move from a gradient of a line between two points to a gradient of a tangent.

### Examples

1. $\lim _{ x\rightarrow 2 }{ \cfrac { { x }^{ 2 }-x-2 }{ x-2 } }$

$\lim _{ x\rightarrow 2 }{ \cfrac { { x }^{ 2 }-x-2 }{ x-2 } } \\ =\lim _{ x\rightarrow 2 }{ \cfrac { (x+1)(x-2) }{ (x-2) } } \\ =\lim _{ x\rightarrow 2 }{ (x+1) } \\ =2+1\\ =3$

2. Find an expression in terms of x for $\lim _{ \delta x\rightarrow 0 }{ \cfrac { 3{ x }^{ 2 }\delta x +\delta { x }^{ 2 }-5\delta x }{ \delta x } }$.

$=\lim _{ \delta x\rightarrow 0 }{ \cfrac { \delta x \left( 3{ x }^{ 2 }+\delta x-5 \right) }{ \delta x } } \\ =\lim _{ \delta x\rightarrow 0 }{ 3{ x }^{ 2 }+\delta x-5 } \\ =3{ x }^{ 2 }-5$

### Differentiation as a Limit

The formula $m=\cfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } }$ is used to find the gradient of a straight line when we know two points on the line. However, when the line is a tangent to a curve, we only know one point on the line-the point of contact with the curve.

To differentiate from first principles, we first use the point of contact and another point close to it on the curve (this line is called a secant) and then we move the second point closer and closer to the point of contact until they overlap and the line is at single point (the tangent). To do this, we use a limit.

### Examples:

1. For the function $f(x)={ x }^{ 3 }$, find the gradient of the secant PQ where P is the point on the function where x=2 and Q is another point on the curve close to P. Choose different values for Q and use these results to estimate the gradient of the curve at P.

We can find a general formula for differentiating from first principles by using c rather than any particular number. We use general points P(c, f(c)) and Q(x, f(x)) where x is close to c.

The gradient of the secant PQ is given by:

$m=\cfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \\ \\ m=\cfrac { f(x)-f(c) }{ x-c }$

The gradient of the tangent at P is found when x approaches c. We call this f'(c).

$f'(c)=\lim _{ x\rightarrow c }{ \cfrac { f(x)-f(c) }{ x-c } }$

There are other versions of this formula.

We can call the points P(x, f(x)) and Q(x+h, f(x+h)) where h is small.

$m=\cfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \\ \\ m=\cfrac { f(x+h)-f(x) }{ x+h-x } \\ \\ m=\cfrac { f(x+h)-f(x) }{ h }$

To find the gradient of the secant, we make h smaller as shown, so that Q becomes closer and closer to P.

As h approaches 0, the gradient of the tangent becomes:

$f^{ ' }\left( x \right) =\lim _{ h\rightarrow 0 }{ \cfrac { f(x+h)-f(x) }{ h } }$

If we use $P(x,y)$ and $Q(x+\delta x,y+\delta y)$ close to P where $\delta x$ and $\delta y$ are small:

$m=\cfrac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } \\ \\ m=\cfrac { y+\delta y-y }{ x+\delta x-x } \\ \\ m=\cfrac { \delta y }{ \delta x }$

As $\delta x$ approaches 0, the gradient of the tangent becomes $\lim _{ \delta x\rightarrow 0 }{ \cfrac { \delta y }{ \delta x } }$, and we call this $\cfrac{dy}{dx}$.

$\cfrac{dy}{dx}=\lim _{ \delta x\rightarrow 0 }{ \cfrac { \delta y }{ \delta x } }$

This notation stands for the derivative, or the gradient of the tangent.

### Examples:

1. Differentiate from first principles to find the gradient of the tangent to the curve $y={x}^{2}+3$ at the point where x=1.

$f^{ ' }\left( x \right) =\lim _{ h\rightarrow 0 }{ \cfrac { f(x+h)-f(x) }{ h } } \\ \\ f(x)={ x }^{ 2 }+3\\ So,\quad f(1)={ 1 }^{ 2 }+3=4\\ \\ f(x+h)={ (x+h) }^{ 2 }+3\\ When\quad x=1\\ f(1+h)={ (1+h) }^{ 2 }+3\\ f(1+h)=1+2h+{ h }^{ 2 }+3\\ So,\quad f(1+h)=2h+{ h }^{ 2 }+4\\ \\ Hence,\\ f^{ ' }\left( x \right) =\lim _{ h\rightarrow 0 }{ \cfrac { f(x+h)-f(x) }{ h } } \\ f^{ ' }\left( 1 \right) =\lim _{ h\rightarrow 0 }{ \cfrac { f(1+h)-f(1) }{ h } } \\ f^{ ' }\left( 1 \right) =\lim _{ h\rightarrow 0 }{ \cfrac { 2h+{ h }^{ 2 }+4-4 }{ h } } \\ f^{ ' }\left( 1 \right) =\lim _{ h\rightarrow 0 }{ \cfrac { h(2+h) }{ h } } \\ f^{ ' }\left( 1 \right) =\lim _{ h\rightarrow 0 }{ \cfrac { (2+h) }{ h } } \\ f^{ ' }\left( 1 \right) =\lim _{ h\rightarrow 0 }{ (2+h) } \\ f^{ ' }\left( 1 \right) =2+0\\ f^{ ' }\left( 1 \right) =2$

2. Differentiate $f\left( x \right) =2{ x }^{ 2 }+7x-3$ from first principles.

$f\left( x \right) =2{ x }^{ 2 }+7x-3\\ \\ f\left( x+h \right) =2{ (x+h) }^{ 2 }+7(x+h)-3\\ f\left( x+h \right) =2({ x }^{ 2 }+2xh+{ h }^{ 2 })+7x+7h-3\\ f\left( x+h \right) =2{ x }^{ 2 }+4xh+2{ h }^{ 2 }+7x+7h-3\\ \\ f\left( x+h \right) -f\left( x \right) =(2{ x }^{ 2 }+4xh+2{ h }^{ 2 }+7x+7h-3)-(2{ x }^{ 2 }+7x-3)\\ f\left( x+h \right) -f\left( x \right) =4xh+2{ h }^{ 2 }+7h\\ \\ f^{ ' }\left( x \right) =\lim _{ h\rightarrow 0 }{ \cfrac { f\left( x+h \right) -f\left( x \right) }{ h } } \\ f^{ ' }\left( x \right) =\lim _{ h\rightarrow 0 }{ \cfrac { 4xh+2{ h }^{ 2 }+7h }{ h } } \\ f^{ ' }\left( x \right) =\lim _{ h\rightarrow 0 }{ \cfrac { h(4x+2{ h }+7) }{ h } } \\ f^{ ' }\left( x \right) =\lim _{ h\rightarrow 0 }{ \quad 4x+2{ h }+7 } \\ f^{ ' }\left( x \right) =\quad 4x+0+7\\ f^{ ' }\left( x \right) =\quad 4x+7$