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- 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
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Tangent and normal: |
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Gradient Function, \(f'(x)\) |
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The function \(f'(x)\) is used to determine the gradient of tangent to the curve at any point on the function graph \(f(x)\).
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- The equation of the tangent is:
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- The line which is perpendicular to the tangent is the normal to the curve \(y=f(x) \text{ pada titik }P(a,f(a))\)
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- The equation of the normal is:
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\(y-f(a)=-\dfrac{1}{f'(a)}(x-a)\) |
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Example |
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Find the equation of the tangent and normal to the curve
\( f(x) = x^3 – 2^2 + 5 \text{ at point } P(2, 5)\).
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Solution:
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\(\text{Given }f(x) = x^3 – 2^2 + 5 , \\ \text{so }f'(x) = 3x^2 -4x\) |
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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 \(P(2,5) \text{ is } -\dfrac{1}{4}.\)
\(\begin{aligned} \text{Equation of the normal is }y – 5 &= -\dfrac{1}{4}(x – 2) \\ 4y – 20 &= -x + 2 \\4y +x &= 22 \end{aligned}\)
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Stationary points of curve \(y=f(x)\) : |
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1) |
Maximum point |
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- Turning points
- When \(x\) increases through \(x=a\), the value of \(\dfrac{dy}{dx}\) changes sign from positive to negative
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2) |
Minimum point |
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- Turning points
- When \(x\) increases through \(x=b\), the value of \(\dfrac{dy}{dx}\) changes sign from negative to positive
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3) |
Point of inflection |
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- 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.
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A turning point on a curve \(y=f(x)\)is a minimum point when |
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\(\dfrac{dy}{dx} = 0 \text{ and } \dfrac{d^2y}{dx^2} \text{ > }0\) |
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A turning point on a curve \(y=f(x)\) is a maximum point when
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\(\dfrac{dy}{dx} = 0 \text{ and } \dfrac{d^2y}{dx^2} \text{ < }0\) |
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Rate of change for related quantities:
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If two variables, \(y\) and \(x\) change with time, \(t\) and are related by the equation \(y=f(x)\), then:
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Chain rule : \(\dfrac{dy}{dt} = \dfrac{dy}{dx} \times \dfrac{dx}{dt}\) |
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Remark |
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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 \(6 \text{ unit }s^{-1}\).
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Small changes and approximations of certain quantities:
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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:
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\(\begin{aligned} \dfrac{\delta y}{\delta x} &\approx \dfrac{dy}{dx}\\\\ \delta y &\approx \dfrac{dy}{dx} \times \delta x\\\\ \text{and }f(x+ \delta x) &\approx y + \delta y\\\\ &\approx y+\dfrac{dy}{dx}(\delta x) \end{aligned}\) |
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Example |
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Given \(y = x^3\), find
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the approximate change in \(y\) when \(x\) increases from \(4\) to \(4.05\),
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(b) |
the approximate change in \(x\) when \(y\) decreases from \(8\) to \(7.97\).
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Solution: |
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(a) |
\(\begin{aligned} y&=x^3\\\\ \dfrac{dy}{dx} &=3x^2 \end{aligned}\)
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When |
\(\begin{aligned} x=4, \delta x &= 4.05-4\\ &=0.05 \end{aligned}\) |
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\(\dfrac{dy}{dx} = 3(4)^2 = 48\) |
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Then, |
\(\begin{aligned} \delta y &\approx \dfrac{dy}{dx} \times \delta x\\\\ &=48 \times 0.05\\\\ \delta y &= 2.4 \end{aligned}\) |
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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 \(2.4\).
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(b) |
When |
\(\begin{aligned} y=8, x^3&=8\\ x&=2\\ \delta y &= 7.97-8 \\ &= -0.03 \end{aligned}\) |
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\(\dfrac{dy}{dx} = 3(2)^2 = 12\) |
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Then, |
\(\begin{aligned} \delta y &\approx \dfrac{dy}{dx} \times \delta x\\\\ -0.03&=12 \times \delta x\\\\ \delta x &= \dfrac{-0.03}{12}\\\\ &=-0.0025 \end{aligned}\) |
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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 \(0.0025\).
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