Differentiation & Basic Integration

Question 6, Question 7 & Question 8

Application of Differentiation # 3• Differentiation can also be used to help us find the equation of a

tangent to a curve at a given point.What is a tangent?

A tangent is a line that touches a curve at one point only!

Application of Differentiation # 3• Hence we can see that a tangent is a straight line.• Can you recall from your study of co – ordinate geometry what

is needed in order to calculate the equation of a line?A point and The slope

• Remember last week when deriving the formula for first principles we found that the slope of the tangent occurs at the first derivative.

Hence if we are given the point of contact of a tangent and a curve then to find the equation of this tangent we:

1. Differentiate the function to find the slope of the tangent.2. Apply the formula for finding the tangent to the curve:

Application of Differentiation # 3

2007 Question 6 (b) (ii)Find the slope of the tangent to at the point (2, ½)

At the moment we have a point on our tangent and so to find the equation of the tangent we will need to find the slope of the tangent also.We know that the slope of the tangent occurs at . Hence we know the slope of the tangent occurs at But how will we get a numerical value for our slope?

We know that the tangent runs through the point (2, ½) and so we can replace the x value in our expression for slope with 2. m = Equation of a line = Tangent: Tangent: 4Tangent: Tangent:

2007 Question 6 (b) (ii)

2004 Question 7 (b) (i)The parametric equations of a curve are

x = y = 1 – cos

Find .

Last week we looked at how we differentiate parametric equations.First we will find = 2 - = 2 - 2cos

Now we will find = 0 - = 2From last week we know that =

2004 Question 7 (b) (i)

Show that the tangent to the curve at is perpendicular to the tanget to the curve at Again we need to rely on our knowledge of co-ordinate geometry to begin this question.If two lines are perpendicular what do we know about their slopes?The slopes of two perpendicular lines when multiplied give an answer of -1.We also know that the slopes of both our tangents will occur at At = =

2004 Question 7 (b) (ii)

At = = = -1 Hence we know the tangent are perpendicular at these points

2004 Question 7 (b) (ii)

Application of Differentiation # 4• Another area where differentiation can help us is in curve

sketching. • We already saw this to some extent last week when we saw how

differentiation can help us to find the turning points of a curve.• Another feature of some (not all graphs) are asymptotes.• The term “asymptote” is a Greek word that literally translates as

“Not touching together”. Hence an asymptote is a line that a curve approaches but never touches.

Application of Differentiation # 4

Vertical Asymptotes• When finding a vertical asymptote we are essentially looking for

a value which x never equal or in more formal terms “for which x is undefined”.

• An instance when x is undefined is when the denominator in a fraction is equal to zero since we cannot divide by zero. Hence to find the vertical asymptote we say:

Vertical Asymptote: Let the denominator equal zero

Horizontal Asymptote• When finding a horizontal asymptote we are essentially looking

for a value which y never equals.

• When finding the horizontal asymptote we look at what value the function, f (x), [or y] is approaching but never touching as x gets bigger and bigger.

• Hence to find the horizontal asymptote we say:

Horizontal Asymptote: y =

2005 Question 6 (c) (i)The equation of the curve is y = where x .Show that the curve has no local maximum or local minimum point

When we get a question like this we will try to find the local max/min points and we will always come across some contradiciton or reach a conclusion that is impossible.Therefore to try to find the turning points we need to find using the Quotient rule.

u = x v = x - 1 1 1

= We know our max and min points occur at = 0 = 0Multiply both sides by the denominator to get -1 = 0But here we have a problem as -1 cannot equal 0 and hence we can conclude that there are no turning points.

2005 Question 6 (c) (i)

Write down the equations of the asymptotes and hence sketch the curve.

To find the vertical asymptote we let the denominator equal zero:x – 1 = 0

To find the horizontal asymptote we must evaluate When we replace x with infinity we get an answer of which we cannot accept.

2005 Question 6 (c) (ii)

2005 Question 6 (c) (ii)

When we get an answer like this we must identify the highest power of x and divide every term in the expression by it.In this instance the highest power of x is x1

=

NB: Any number divided by infinity is zero. = = 1Horizontal Asymptote: y = 1

2005 Question 6 (c) (ii)

Application of Differentiation # 5• The final application of differentiation that we will look at is the

way in which it helps us to approximate roots of equations.• We are used to finding roots of quadratic equations i.e. those of

the form ax2 + bx + c but in fact differentiation can help us approximate roots of much more complex equations.

• The way we can do this is through the Newton Raphson method!

Newton Raphson Method• The Newton Raphson method is based on the premise that if we

are looking to approximate roots and we approximate xn then xn+1 is a better approximation where:

• Hence to find the (n + 1)st approximation we must: 1. Sub xn into our function 2. Differentiate our function 3. Sub xn into our differentiated function 4. Sub our three values into the Newton Method formula.

𝑥𝑛+1=𝑥𝑛−𝑓 (𝑥𝑛)𝑓 ′ (𝑥𝑛)

2006 Question 7(a)Taking x1 = 2 as the first approximation to the real root of the equation x3 + x – 9 = 0, use the Newton Raphson method to find x2 the second approximation.

f (x) = x3 + x – 9 f (x1) = (2)3 + (2) – 9 = 1f (x) = 3x2 + 1f (x1) = 3(2)2 + 1 = 13 =

2003 Question 6 (b)Show that the equation x3 – 4x – 2 = 0 has a root between 2 and 3.Taking x1 = 2 as the first approximation to this root use the Newton Raphson method to find x3 the third approximation.If a root lies between two values a and b then the sign of f (a) and f (b) will be different. Why? Because in order for a root to lie between these two values it means that some where between these two values our function crosses the x axis and the function goes from positive to negative or vice versa.f (x) = x3 – 4x – 2f (2) = 23 – 4(2) – 2 = - 2f (3) = 33 – 4(3) – 2 = 13Since the sign of f (2) is different to that of f (3) we know a root lies between 2 and 3.

f (x) = x3 – 4x – 2 and x1 = 2f (x1) = (2)3 - 4(2) – 2 = -2f (x) = 3x2 -4f (x1) = 3(2)2 -4= 8 = f (x2) = ()3 - 4() – 2 = f (x) = 3x2 -4f (x1) = 3()2 -4=

2003 Question 6 (b)

2003 Question 6 (b)

Implicit Differentiation• To date we have looked at differentiating functions of the form

y = f (x) were we have y on one side and all expressions containing x on the other side.

• However all functions are not written in this explicit manner with the x’s and y’s split up.

• Instead some times we have to deal with implicit functions. An implicit function is one that implies a function exists i.e. a direct relationship between x and y exists, without ever stating this function.

• An example of an implicit function might be: 2x2 + 5xy – y2

Implicit Differentiation• In order to differentiate such functions we must differentiate

both sides of our equation with respect to x and then look to isolate .

• We will rely heavily on the Chain and product rule when doing this as now when we have an expression such as xy we will need to treat these as the product of variables.

• A key thing to remember and something which is derived from the chain rule is that if you are differentiating y w.r.t. x then you differentiate y as normal and then multiply by . For example if you were to differentiate y2 w.r.t. x then the derivative would be 2y

2005 Question 7(b) (ii)Find the slope of the tangent to the curve xy2 + y =6 at the point (1,2)

We know the slope of the tangent to the curve occurs at To find we must differentiate every term with respect to x.Derivative of y = 1 Derivative of 6 = 0To differentiate xy2 we must use the chain rule:u = x v = y2

= 2y

Derivative of xy2 = (x)(2y) + (y2)(1) = 2xy + y2 2xy + y2 + = 02xy + = - y2 Take out the common factor of on the LHS(2xy + 1) = - y2 To find the slope we need to evaluate this at the point (1, 2)

2005 Question 7(b) (ii)

2005 Question 7(b) (ii)

Introduction to Integration• Integration is the opposite of differentiation. When we studied

differentiation we saw how differentiation allowed us to find the rate of change of one variable with respect to another. Since integration is the opposite process it allows us to find an expression for a variable given its rate of change with respect to another variable.

• Integration was developed along with differentiation and some of it’s earliest uses was to find the area of irregular shapes e.g. lakes and to find the volume of objects.

• It is still used today by electricians measuring electric charges or even by car manufacturers when ensuring that there car meets the required safety standards.

Relationship between Differentiation and Integration

f(x)

Differentiate

f '(x)

Integrate

• The last slide showed how differentiation and integration are opposite processes.

• If you remember when we were asked to differentiate we multiply by the power and then reduce the power by 1. Since integration is the opposite process we do the opposite procedures in the opposite order. That means we increase the power by 1 and then divide by the new power.

• However there is one thing we need to understand when integrating.

Relationship between Differentiation and Integration

Constant of Integration

If we are given the function f (t) = 2t 3 – 4t 2 + 7t – 4 we know if we differentiate it we get: Hence we know if we integrate this function we should get back to our original function: = 2t 3 – 4t 2 + 7tIs this the same as the function we started with? What is different?

Constant of Integration• In the integrated function we are missing the number/constant (-

4) from our original equation.• The reason for this is because when we differentiate the

derivative of the constant is zero. • As a result it is clear that when we integrate a function the

constant, if there was originally one, is omitted.• We must counteract this by always adding a constant C when

integrating to ensure we get the correct answer (Note: If there is no constant in the original function then C = 0)

2008 Question 8 (a)

2007 Question 8 (a)