Applications of Derivatives in IB Mathematics

Applications of Derivatives in IB

Mathematics-

In my previous post, we discussed how to find the derivative of different types of functions as well as the geometrical meaning of differentiation. Here we are discussing  Applications of Derivatives in IB Mathematics
There are many different fields for the Applications of Derivatives. We shall discuss a few of them-

Slope and Equation of tangents to a curve- If We draw a tangent to a curve y=f(x) at a given point   ({x_1},{y_1}), then

The gradient of the curve at given point=the gradient of the tangent line  at given  point

and we already discussed that slope or gradient of the tangent at given point   ({x_1},{y_1})

m=  {\frac{{dy}}{{dx}}_{at({x_1},{y_1})}}

=f'({x_1})

Finally to find the equation of tangent we use the slope-point form of equation

y - {y_1} = m(x - {x_1})

The major part of this concept is also discussed in the previous post. We should also remember following points while solving these types of questions.

(i) If two lines are parallel to each other, their slopes are always equal
i.e     {m_1} = {m_2}
(ii) If two lines are perpendicular to each other, the product of their  slopes is always -1

{m_1}.{m_2} = - 1

(iii) If a line is passing through two points   ({x_1},{y_1}) and  ({x_2},{y_2})  then, slope of the line

m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}}

Example:- This is the equation of a curve  3x - 2{y^2}{e^{x - 1}} = 2

(a) Find  \frac{{dy}}{{dx}} in terms of x and y

(b) Find the equations of the tangents to this curve at the points where the curve intersects the line at x = 1.

Ans:

(a) We are given an implicit function we can differentiate both sides with respect to x

                      \frac{d}{{dx}}[3x - 2{y^2}{e^{x - 1}} = 2]

                                      \begin{array}{l} 3 - 2\frac{d}{{dx}}({y^2}{e^{x - 1}}) = 0\\ 3 = 2\left( {2y{e^{x - 1}}\frac{{dy}}{{dx}} + {e^{x - 1}}.{y^2}} \right) \end{array}

                                    3 - 2{e^{x - 1}}.{y^2} = 4y{e^{x - 1}}\frac{{dy}}{{dx}}

                  \frac{{dy}}{{dx}} = \frac{{3 - 2{e^{x - 1}}.{y^2}}}{{4y{e^{x - 1}}}}

 

(b)  let x=1 in the given equation of the curve, we get   y = \pm \sqrt {\frac{1}{2}}

so the point of contacts is  \left( {1,\sqrt {\frac{1}{2}} } \right)   or   \;\left( {1, - \sqrt {\frac{1}{2}} } \right)

and slope of tangent m= \frac{{dy}}{{dx}}{\;_{at({x_1},{y_1})}}

=   \frac{{3 - 2.{{\left( {\sqrt {\frac{1}{2}} } \right)}^2}}}{{4\left( { \pm \sqrt {\frac{1}{2}} } \right)}}

=      \frac{{3 - 1}}{{ \pm 2\sqrt 2 }}

m  =      \pm \frac{1}{{\sqrt 2 }}

now we have slope and point of contact, we can put these values in the following formula

 

y - {y_1} = m(x - {x_1})

final equation of line will be    y = \pm \frac{1}{{\sqrt 2 }}x

This was an example of Applications of Derivatives in IB Mathematics. It was asked in Nov.,2016 IB Mathematics HL exam

Slope and Equation of Normal to a Curve- Normal is a line which is perpendicular to the curve at the point of contact. If we write slope of a tangent as  {m_T} and slope of normal as  m{}_N

then                                            m{}_N.{m_T} =-1

so  m{}_N=-1/{m_T}

the equation of normal                            \Rightarrow  \begin{array}{lllllllllllllll} \;\\ {y - {y_1} = \frac{{ - 1}}{{{m_T}}}(x - {x_1})} \end{array}

The point P(x1 , y1) will satisfy the equation of the curve & the equation of tangent &      normal line.

► The point P(x1 , y1) will satisfy the equation of the curve & the equation of tangent & normal line.v If the tangent at any point P on the curve is parallel to the axis of x then

If the tangent at any point P on the curve is parallel to the axis of x then dy/dx = 0 at the point P.

If the tangent at any point on the curve is parallel to the axis of y, then dy/dx = \infty or dx/dy = 0.

If the tangent at any point on the curve is equally inclined to both the axes then dy/dx = ± 1.

If the tangent at any point makes equal intercept on the coordinate axes then dy/dx =-1.

Tangent to a curve at the point P (x1 , y1)  can be drawn even through dy/dx at P does not exist. e.g.  x = 0 is a tangent to  y = x2/3  at (0, 0).

If a curve passing through the origin be given by a rational integral algebraic equation, the equation of the tangent (or tangents) at the origin is obtained by equating to zero the terms of the lowest degree in the equation. e.g. If the equation of a curve be

x2 – y2 + x3 + 3 x2 y – y3 = 0 , the tangents at the origin are given by x2 – y2= 0 i.e.

                              x + y = 0 and x – y = 0.

The angle of intersection between two curves- Angle of intersection between two curves is defined as the angle between the 2 tangents drawn to the 2 curves at their point of intersection. If the angle between two curves is 90° everywhere then they are called Orthogonal curves. If slope of tangents be  {m_1} and {m_2}  the angle between the two curves is  \theta  then

                                                  \tan \theta = \frac{{\left| {{m_1} - {m_2}} \right|}}{{1 + {m_1}{m_2}}}

These are a few fields of  Applications of Derivatives in IB Mathematics.
In the next post, we shall discuss Maxima and Minima, that’s also a very useful example of  Applications of Derivatives in IB Mathematics

Click here to download worksheet of tangent and normal question

 Worksheets on Tangent Normal (22-09-15).pdf

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