In this note, you will learn:

1) Differentiation with relation to equation of tangent and normal to curve

2) Differentiation with relation to connected rate of change

### 1) Differentiation with relation to equation of tangent and normal to curve

Previously on the Differentiation (Part I) and Differentiation (Part II), we have learnt the definition of differentiation and the different rules of differentiation. Today, we will be learning about differentiation with relation to equation of tangent and normal to curve.

From the figure shown above, you can see that it shows the curve, y = f(x), where l1 is the tangent to the curve at the point (x1, y1).

The gradient of the tangent to the curve at any point x is given by the derivative of the function, which is dy/dx.

If it is given that the value of dy/dx at (x1, y1) is m, the equation of the tangent will be as follows:

y â€“ y1 = m (x â€“ x1)

This formula is also known as the point-slope formula, and is used when the y-intercept is not known but the gradient of the line is given. The perk of using this point-slope formula is that you do not need to know two points on the line, which you normally do in order to find the gradient using the formula: gradient = (y1 â€“ y2)/ (x1 â€“ x2).

The line l2 is perpendicular to the tangent and is known as the normal to the curve at (x1, y1).

If m â‰ 0, the gradient of l2 is given by -1/m and its equation will be as follows:

y â€“ y1 = (-1/m) (x â€“ x1)

Letâ€™s take a look at how these work in the example shown below:

Question: Find the equations of the tangent and the normal to the curve y = 3x^3 â€“ 5x^2 + 9 at the point where x = 2.

Solution:

y = 3x^3 â€“ 5x^2 + 9

dy/dx = 9x^2 â€“ 10x

When x = 4, y = 3(2)^3 â€“ 5(2)^2 + 9 = 13

and dy/dx = 9(2)^2 â€“ 10(2) = 16

i.e. at (2, 13), gradient of tangent = 16

Therefore, equation of tangent is y â€“ 13 = 16(x â€“ 2)

y = 16x â€“ 19

Gradient of normal at (2, 13) = -1/16

Therefore, equation of normal is y â€“ 13 = (-1/16) (x- 2)

y = (-1/16) x + 105/8

### Differentiation with relation to connected rate of change

Now, there is another way in which differentiation can be applied: Rate of change.

Given that we have a variable, x, and it is said that x is changing 4 times as fast as another variable, v, and v is changing 5 times as fast as variable u. So, how fast is x changing with respect to u? Based on our visual calculations, we can see that x is changing 20 times as fast as u.

This is very relatable to differentiation, because what we are essentially calculating when we that the derivative of a function is the difference or the change of a variable with respect to the change of another variable! The larger the difference, means that the greater the rate of change with respect to the other variable.

Examples for you to visualize can be seen below:

Â· dy/dx (rate of change of y with respect to x)

Â· dy/du (rate of change of y with respect to u)

Â· du/dx (rate of change of u with respect to x)

These derivatives might seem familiar to you, and remind you of something we have learnt previously in the first part of differentiation: Chain Rule!

Such an application is relevant to the problems of our everyday life. For example, if we fill a cuboid tank with water from a running tap at a constant rate of 20ml/s. Given that the base area of the cuboid tank is 35cm2, how fast is the height of the water level in the tank increasing 10 seconds after the tap was switched on?

We must hence consider the relationship between the height of the water and the volume of the tank to relate the rate of change of the height of the water and the rate of change of the volume of the tank. These are called connected rate of change/ related rates problems.

As mentioned above, the rate of change of the volume, dV/dt, is 20ml/s. What we are trying to find is the rate of change of the height of the water level in the tank, dh/dt. So, how do we do it?

The formula for the volume of a cuboid is Volume = length x base x height, and since we are already provided with the base area (length x base), we can simplify the equation to be:

Volume = base area x height

Or: V = 35h

Hence, we can find the rate of change of the volume of the tank with respect to its height.

Using our knowledge of chain rule, we can form the equation: dV/dt = (dV/dh) x (dh/dt)

Since we know the values of dV/dt and dV/dh, we can now find the value of dh/dt.

Letâ€™s take a look at some of the examples shown below regarding the application of connected rate of change:

Q1) Given that M = (1/4) Ï€x^4 + 3 Ï€x^2 + 8x, find the rate of change of M with respect to x when x = 3, leaving your answer in terms of Ï€.

Solution:

M = (1/4) Ï€x^4 + 3 Ï€x^2 + 8x

dM/dx = (1/4) Ï€(4x^3) + 3Ï€(2x) + 8

= Ï€x3 + 6Ï€x + 8

When x = 3, dV/dx = Ï€ (3)^3 + 6Ï€ (3) + 8

= 27Ï€ + 18Ï€ + 8

= 45Ï€ + 8

Q2) The radius of a circle increases at a rate of 0.5cm/s. Calculate the rate of increase in the area when the radius is 8 cm, leaving your answer in terms of Ï€.

Solution:

Let r cm and A cm^2 be the radius and the area of the circle respectively, let t s be the time.

Then dr/dt = 0.5.

A = Ï€r^2

dA/dr = 2Ï€r

dA/dt = (dA/dr) x (dr/dt)

= 2Ï€r x 0.5

= Ï€r

When r = 8, dA/dt = Ï€ x 8

= 8Ï€

Therefore, the rate of increase in the area is 8Ï€ cm^2/s.

Try it yourself!

A hemispherical bowl of radius 6 cm is completely filled with water. The water is then transferred at a steady rate into an empty inverted right circular cone of base radius 6 cm and height 14 cm.

Given that all the water is transferred in 16/3 seconds and that V cm^3 represents the amount of water in the cone at any time t, find

i) dV/dt in terms of Ï€,

ii) the rate of change of

a) the height of the water level,

b) the horizontal surface area of the cone

when the depth of the water is 6 cm.

And thatâ€™s all for today, students! Math Lobby hopes that after this article, you have a clear understanding on the topic of the application of differentiation with relations to equation of tangent and normal, and connected rate of change!

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