Determining Curve Equations Using Integrals
Smaridasa Mathematics – From the definition of the integral, we obtained the relationship between derivatives and integrals. In this article, we will discuss the topic of Determining Curve Equations Using Integrals, where we will determine the equation of a curve from the derivative of that curve. The prerequisite material that must be mastered first is "Indefinite Integral of Algebraic Functions", as here we will only discuss algebraic function forms. Actually, we can find the equation of any type of function if we know its derivative.
Concept of Determining Curve Equations with Integrals
From the definition of the integral, we have \( \int f(x) \, dx = F(x) + c \) where the derivative of \( (F(x) + c) \) is \( f(x) \), or we can write \( F'(x) = f(x) \). If we replace \( f(x) = F'(x) \), then we get:
\[ \int f(x) \, dx = F(x) + c \quad \Leftrightarrow \quad \int F'(x) \, dx = F(x) + c. \]
From the form \( \int F'(x) \, dx = F(x) + c \), it means that to remove the derivative, we simply integrate it. Another way to write this is:
\[ f(x) = \int f'(x) \, dx, \quad \text{or} \quad y = \int y' \, dx, \quad \text{or} \quad S(t) = \int S'(t) \, dt. \]
Important Notes:
1. To remove a derivative, simply integrate the derivative function.
2. If the second derivative is known, integrate twice to obtain the original curve function. Continue this process, integrating as many times as the known derivative order.
3. The result of integrating the derivative is called a family of curves because it still contains the constant \( +c \). The constant \( c \) can take various values, generating many possible curve equations.
4. To determine a specific value of \( c \), substitute a point that lies on the curve.
Example Problems: Determining Curve Equations
1. A curve passes through the point (2, 1). If the gradient of the curve at any point satisfies the relation:
\[ \frac{dy}{dx} = 2\left( x - \frac{1}{x^2} \right), \]
determine the equation of the curve.
Solution:
Given the derivative: \( y' = 2\left( x - \frac{1}{x^2} \right) \).
Integrate the derivative:
\[ y = \int y' \, dx = \int 2\left( x - \frac{1}{x^2} \right) dx \]
\[ = \int 2\left( x - x^{-2} \right) dx = 2 \int \left( x - x^{-2} \right) dx \]
\[ = 2 \left( \frac{1}{2}x^2 - \frac{1}{-1}x^{-1} \right) + c = 2 \left( \frac{1}{2}x^2 + \frac{1}{x} \right) + c \]
\[ y = x^2 + \frac{2}{x} + c \]
Substitute the point (2, 1) to find \( c \):
\[ 1 = 2^2 + \frac{2}{2} + c \quad \Rightarrow \quad 1 = 4 + 1 + c \quad \Rightarrow \quad c = -4 \]
Thus, the equation of the curve is:
\[ y = x^2 + \frac{2}{x} - 4. \]
2. If a curve \( F(x) \) passes through the point (1, 3) with \( F'(x) = 3x^2 + 4x - 1 \), determine the value of \( F(-1) \).
Solution:
First find \( F(x) \) by integrating:
\[ F(x) = \int F'(x) \, dx = \int (3x^2 + 4x - 1) dx = x^3 + 2x^2 - x + c \]
Use the point (1, 3) to find \( c \):
\[ 3 = 1^3 + 2(1)^2 - 1 + c \quad \Rightarrow \quad 3 = 2 + c \quad \Rightarrow \quad c = 1 \]
Thus, \( F(x) = x^3 + 2x^2 - x + 1 \).
Now find \( F(-1) \):
\[ F(-1) = (-1)^3 + 2(-1)^2 - (-1) + 1 = -1 + 2 + 1 + 1 = 3 \]
3. The marginal cost (MC) of producing a certain good (Q) each month is given by \( MC = \frac{dC}{dQ} = 2Q + 3 \). The cost to produce 1 unit is 300 (in thousands of rupiah). Determine the total cost function per month.
Note: Q = quantity, C = total cost, MC = marginal cost.
Solution:
Given: \( C'(Q) = 2Q + 3 \).
Integrate to find \( C(Q) \):
\[ C(Q) = \int (2Q + 3) dQ = Q^2 + 3Q + k \]
Given \( C(1) = 300 \) (in thousands):
\[ 300 = 1^2 + 3(1) + k \quad \Rightarrow \quad 300 = 4 + k \quad \Rightarrow \quad k = 296 \]
Thus, the total cost function is:
\[ C(Q) = Q^2 + 3Q + 296 \quad \text{(in thousands of rupiah)}. \]
4. Determine the function \( y = f(x) \) from the differential equation \( \frac{x^2 \, dy}{dx} = y^2 \sqrt{x} \) with \( y = 1 \) when \( x = 1 \).
Solution:
Separate variables and integrate:
\[ \frac{x^2 \, dy}{dx} = y^2 \sqrt{x} \quad \Rightarrow \quad \frac{dy}{y^2} = \frac{\sqrt{x}}{x^2} dx \]
\[ \int y^{-2} dy = \int x^{-3/2} dx \]
\[ -y^{-1} = -2 x^{-1/2} + c \quad \Rightarrow \quad \frac{1}{y} = \frac{2}{\sqrt{x}} + c \]
Use \( y=1, x=1 \):
\[ 1 = \frac{2}{\sqrt{1}} + c \quad \Rightarrow \quad c = -1 \]
Thus:
\[ \frac{1}{y} = \frac{2}{\sqrt{x}} - 1 \quad \Rightarrow \quad y = \frac{\sqrt{x}}{2 - \sqrt{x}}. \]
5. A curve \( y = f(x) \) passes through (1, 2) with a gradient of -5 at that point. If \( f''(x) = 6x + 4 \), find the equation of the curve.
Solution:
First integrate \( f''(x) \) to get \( f'(x) \):
\[ f'(x) = \int (6x + 4) dx = 3x^2 + 4x + c_1 \]
Given \( f'(1) = -5 \):
\[ -5 = 3(1)^2 + 4(1) + c_1 \quad \Rightarrow \quad c_1 = -12 \]
So \( f'(x) = 3x^2 + 4x - 12 \).
Integrate again to find \( f(x) \):
\[ f(x) = \int (3x^2 + 4x - 12) dx = x^3 + 2x^2 - 12x + c_2 \]
Given \( f(1) = 2 \):
\[ 2 = 1^3 + 2(1)^2 - 12(1) + c_2 \quad \Rightarrow \quad c_2 = 11 \]
Thus, the curve equation is:
\[ f(x) = x^3 + 2x^2 - 12x + 11. \]
Application in Physics
In physics, we have position \( s(t) \), velocity \( v(t) \), and acceleration \( a(t) \) functions. From kinematics:
\[ v(t) = s'(t) \quad \Rightarrow \quad s(t) = \int v(t) \, dt \]
\[ a(t) = v'(t) \quad \Rightarrow \quad v(t) = \int a(t) \, dt \]
Physics Example
6. A particle's acceleration is given by \( a(t) = -2t^2 + 3t + 1 \). Determine the position function \( s(t) \) if the initial velocity is \( v_0 = 2 \) and initial position is \( s_0 = 1 \).
Solution:
Find velocity by integrating acceleration:
\[ v(t) = \int a(t) \, dt = \int (-2t^2 + 3t + 1) dt = -\frac{2}{3}t^3 + \frac{3}{2}t^2 + t + C_1 \]
Given \( v(0) = 2 \):
\[ 2 = 0 + 0 + 0 + C_1 \quad \Rightarrow \quad C_1 = 2 \]
Thus \( v(t) = -\frac{2}{3}t^3 + \frac{3}{2}t^2 + t + 2 \).
Find position by integrating velocity:
\[ s(t) = \int v(t) \, dt = \int \left( -\frac{2}{3}t^3 + \frac{3}{2}t^2 + t + 2 \right) dt \]
\[ = -\frac{1}{6}t^4 + \frac{1}{2}t^3 + \frac{1}{2}t^2 + 2t + C_2 \]
Given \( s(0) = 1 \):
\[ 1 = 0 + 0 + 0 + 0 + C_2 \quad \Rightarrow \quad C_2 = 1 \]
Thus the position function is:
\[ s(t) = -\frac{1}{6}t^4 + \frac{1}{2}t^3 + \frac{1}{2}t^2 + 2t + 1. \]
General Formula
\[ f(x) = \int f'(x) \, dx \]