## Application of Differentiation

 2.4 Application of Differentiation

• Safety and maximum enjoyment of customers
• The building of a roller coaster also take into consideration on users' maximum enjoyment out of the ride
• The gradient of a curve at a point is also the gradient of the tangent at that point

Tangent and normal:

Gradient Function, $$f'(x)$$
 The function $$f'(x)$$ is used to determine the gradient of tangent to the curve at any point on the function graph $$f(x)$$.

• Based on the above diagram, the formula for the equation of straight line $$l$$ with gradient $$m$$ that passes through point $$P(x_1, y_1)$$ can be written as:

 $$y-y_1=m(x-x_1)$$

• The equation of the tangent is:
 $$y-f(a)=f'(a)(x-a)$$

• The line which is perpendicular to the tangent is the normal to the curve  $$y=f(x) \text{ pada titik }P(a,f(a))$$

• If the gradient of the tangent, $$f'(a)$$ exists and is non-zero, the gradient of the normal based on the relation of $$m_1m_2=-1$$ is

 $$-\dfrac{1}{f'(a)}$$

• The equation of the normal is:
 $$y-f(a)=-\dfrac{1}{f'(a)}(x-a)$$

 Example Find the equation of the tangent and normal to the curve $$f(x) = x^3 – 2^2 + 5 \text{ at point } P(2, 5)$$. Solution: $$\text{Given }f(x) = x^3 – 2^2 + 5 , \\ \text{so }f'(x) = 3x^2 -4x$$ When $$x=2,\ f'(2) \ = \ 3(2)^2 -4(2) = \ 12-8 \ = \ 4$$   Gradient of the tangent at point $$P(2,5) \text{ is } 4.$$   \begin{aligned} \text{Equation of the tangent is }y – 5 &= 4(x – 2) \\ y – 5 &= 4x – 8 \\&y = 4x – 3 \end{aligned}   Gradient of the normal at point $$P(2,5) \text{ is } -\dfrac{1}{4}.$$   \begin{aligned} \text{Equation of the normal is }y – 5 &= -\dfrac{1}{4}(x – 2) \\ 4y – 20 &= -x + 2 \\4y +x &= 22 \end{aligned}

Stationary points of curve $$y=f(x)$$ :

• There are three types of stationary points:

 1) Maximum point Turning points When $$x$$ increases through $$x=a$$, the value of $$\dfrac{dy}{dx}$$ changes sign from positive to negative 2) Minimum point Turning points When $$x$$ increases through $$x=b$$, the value of $$\dfrac{dy}{dx}$$ changes sign from negative to positive 3) Point of inflection This stationary point which is not a maximum or a minimum point A point on the curve at which the curvature of the graph changes.

• In general,
 A turning point on a curve $$y=f(x)$$is a minimum point when $$\dfrac{dy}{dx} = 0 \text{ and } \dfrac{d^2y}{dx^2} \text{ > }0$$

 A turning point on a curve $$y=f(x)$$ is a maximum point when $$\dfrac{dy}{dx} = 0 \text{ and } \dfrac{d^2y}{dx^2} \text{ < }0$$

Rate of change for related quantities:

• If two variables, $$y$$ and $$x$$ change with time, $$t$$ and are related by the equation $$y=f(x)$$, then:

 Chain rule : $$\dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt}$$

Remark
 If the rate of change of y over time is negative, for example $$\dfrac{dy}{dt} = -6$$, then  $$y$$ is said to decrease at a rate of $$6 \text{ unit }s^{-1}$$.

Small changes and approximations of certain quantities:

• If $$y=f(x)$$ and the small change in $$x$$, that is $$\delta x$$ causes a small change in $$y$$, that is $$\delta y$$, then:

 \begin{aligned} \dfrac{\delta y}{\delta x} &\approx \dfrac{dy}{dx}\\\\ \delta y &\approx \dfrac{dy}{dx} \times \delta x\\\\ \text{and }f(x+ \delta x) &\approx y + \delta y\\\\ &\approx y+\dfrac{dy}{dx}(\delta x) \end{aligned}

Example

Given $$y = x^3$$, find

(a)

the approximate change in $$y$$ when $$x$$ increases from $$4$$ to $$4.05$$,

(b)

the approximate change in $$x$$ when $$y$$ decreases from $$8$$ to $$7.97$$.

Solution:

(a)

\begin{aligned} y&=x^3\\\\ \dfrac{dy}{dx} &=3x^2 \end{aligned}

 When \begin{aligned} x=4, \delta x &= 4.05-4\\ &=0.05 \end{aligned} and $$\dfrac{dy}{dx} = 3(4)^2 = 48$$ Then, \begin{aligned} \delta y &\approx \dfrac{dy}{dx} \times \delta x\\\\ &=48 \times 0.05\\\\ \delta y &= 2.4 \end{aligned}

Therefore, the approximate change in $$y$$, that is $$\delta y$$, is $$2.4$$.

$$\delta y \text{ > }0$$ means there is a small increase in $$y$$ of $$2.4$$.

(b)
 When \begin{aligned} y=8, x^3&=8\\ x&=2\\ \delta y &= 7.97-8 \\ &= -0.03 \end{aligned} and $$\dfrac{dy}{dx} = 3(2)^2 = 12$$ Then, \begin{aligned} \delta y &\approx \dfrac{dy}{dx} \times \delta x\\\\ -0.03&=12 \times \delta x\\\\ \delta x &= \dfrac{-0.03}{12}\\\\ &=-0.0025 \end{aligned}

Therefore, the approximate change in $$x$$, that is $$\delta x$$, is $$-0.0025$$.

$$\delta y \text{ < }0$$ means there is a small decrease in $$x$$ of $$0.0025$$.

## Application of Differentiation

 2.4 Application of Differentiation

• Safety and maximum enjoyment of customers
• The building of a roller coaster also take into consideration on users' maximum enjoyment out of the ride
• The gradient of a curve at a point is also the gradient of the tangent at that point

Tangent and normal:

Gradient Function, $$f'(x)$$
 The function $$f'(x)$$ is used to determine the gradient of tangent to the curve at any point on the function graph $$f(x)$$.

• Based on the above diagram, the formula for the equation of straight line $$l$$ with gradient $$m$$ that passes through point $$P(x_1, y_1)$$ can be written as:

 $$y-y_1=m(x-x_1)$$

• The equation of the tangent is:
 $$y-f(a)=f'(a)(x-a)$$

• The line which is perpendicular to the tangent is the normal to the curve  $$y=f(x) \text{ pada titik }P(a,f(a))$$

• If the gradient of the tangent, $$f'(a)$$ exists and is non-zero, the gradient of the normal based on the relation of $$m_1m_2=-1$$ is

 $$-\dfrac{1}{f'(a)}$$

• The equation of the normal is:
 $$y-f(a)=-\dfrac{1}{f'(a)}(x-a)$$

 Example Find the equation of the tangent and normal to the curve $$f(x) = x^3 – 2^2 + 5 \text{ at point } P(2, 5)$$. Solution: $$\text{Given }f(x) = x^3 – 2^2 + 5 , \\ \text{so }f'(x) = 3x^2 -4x$$ When $$x=2,\ f'(2) \ = \ 3(2)^2 -4(2) = \ 12-8 \ = \ 4$$   Gradient of the tangent at point $$P(2,5) \text{ is } 4.$$   \begin{aligned} \text{Equation of the tangent is }y – 5 &= 4(x – 2) \\ y – 5 &= 4x – 8 \\&y = 4x – 3 \end{aligned}   Gradient of the normal at point $$P(2,5) \text{ is } -\dfrac{1}{4}.$$   \begin{aligned} \text{Equation of the normal is }y – 5 &= -\dfrac{1}{4}(x – 2) \\ 4y – 20 &= -x + 2 \\4y +x &= 22 \end{aligned}

Stationary points of curve $$y=f(x)$$ :

• There are three types of stationary points:

 1) Maximum point Turning points When $$x$$ increases through $$x=a$$, the value of $$\dfrac{dy}{dx}$$ changes sign from positive to negative 2) Minimum point Turning points When $$x$$ increases through $$x=b$$, the value of $$\dfrac{dy}{dx}$$ changes sign from negative to positive 3) Point of inflection This stationary point which is not a maximum or a minimum point A point on the curve at which the curvature of the graph changes.

• In general,
 A turning point on a curve $$y=f(x)$$is a minimum point when $$\dfrac{dy}{dx} = 0 \text{ and } \dfrac{d^2y}{dx^2} \text{ > }0$$

 A turning point on a curve $$y=f(x)$$ is a maximum point when $$\dfrac{dy}{dx} = 0 \text{ and } \dfrac{d^2y}{dx^2} \text{ < }0$$

Rate of change for related quantities:

• If two variables, $$y$$ and $$x$$ change with time, $$t$$ and are related by the equation $$y=f(x)$$, then:

 Chain rule : $$\dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt}$$

Remark
 If the rate of change of y over time is negative, for example $$\dfrac{dy}{dt} = -6$$, then  $$y$$ is said to decrease at a rate of $$6 \text{ unit }s^{-1}$$.

Small changes and approximations of certain quantities:

• If $$y=f(x)$$ and the small change in $$x$$, that is $$\delta x$$ causes a small change in $$y$$, that is $$\delta y$$, then:

 \begin{aligned} \dfrac{\delta y}{\delta x} &\approx \dfrac{dy}{dx}\\\\ \delta y &\approx \dfrac{dy}{dx} \times \delta x\\\\ \text{and }f(x+ \delta x) &\approx y + \delta y\\\\ &\approx y+\dfrac{dy}{dx}(\delta x) \end{aligned}

Example

Given $$y = x^3$$, find

(a)

the approximate change in $$y$$ when $$x$$ increases from $$4$$ to $$4.05$$,

(b)

the approximate change in $$x$$ when $$y$$ decreases from $$8$$ to $$7.97$$.

Solution:

(a)

\begin{aligned} y&=x^3\\\\ \dfrac{dy}{dx} &=3x^2 \end{aligned}

 When \begin{aligned} x=4, \delta x &= 4.05-4\\ &=0.05 \end{aligned} and $$\dfrac{dy}{dx} = 3(4)^2 = 48$$ Then, \begin{aligned} \delta y &\approx \dfrac{dy}{dx} \times \delta x\\\\ &=48 \times 0.05\\\\ \delta y &= 2.4 \end{aligned}

Therefore, the approximate change in $$y$$, that is $$\delta y$$, is $$2.4$$.

$$\delta y \text{ > }0$$ means there is a small increase in $$y$$ of $$2.4$$.

(b)
 When \begin{aligned} y=8, x^3&=8\\ x&=2\\ \delta y &= 7.97-8 \\ &= -0.03 \end{aligned} and $$\dfrac{dy}{dx} = 3(2)^2 = 12$$ Then, \begin{aligned} \delta y &\approx \dfrac{dy}{dx} \times \delta x\\\\ -0.03&=12 \times \delta x\\\\ \delta x &= \dfrac{-0.03}{12}\\\\ &=-0.0025 \end{aligned}

Therefore, the approximate change in $$x$$, that is $$\delta x$$, is $$-0.0025$$.

$$\delta y \text{ < }0$$ means there is a small decrease in $$x$$ of $$0.0025$$.