# Thread: [SOLVED] Differentiation - Equation of curve

1. ## [SOLVED] Differentiation - Equation of curve

Question:
A curve is such that dy/dx = 2x²⁪ − 5. Given that the point (3, 8) lies on the curve, find the equation of the curve.

Attempt:
dy/dx = 2x²⁪-5
x = 3
2(3)²-5 = 13

y - y1 = m (x - x1)
y - 8 = 13 (x - 3)
y - 8 = 13x - 39
y = 13x - 39 +8
y = 13x -31

2. Originally Posted by looi76
Question:
A curve is such that dy/dx = 2x²⁪ − 5. Given that the point (3, 8) lies on the curve, find the equation of the curve.

Attempt:
dy/dx = 2x²⁪-5
x = 3
2(3)²-5 = 13

y - y1 = m (x - x1)
y - 8 = 13 (x - 3)
y - 8 = 13x - 39
y = 13x - 39 +8
y = 13x -31
(3,8) lies on the initial curve, you have to integrate/anti-differentiate in order to find the formula for this.

$\frac{dy}{dx} = 2x^2-5$

$y = \frac{2}3x^3-5x+C$

Now you only need to determine the value of your constant, which can be done by plugging in your x and y values.

3. Originally Posted by looi76
Question:
A curve is such that dy/dx = 2x²⁪ − 5. Given that the point (3, 8) lies on the curve, find the equation of the curve.

Attempt:
dy/dx = 2x²⁪-5
x = 3
2(3)²-5 = 13

y - y1 = m (x - x1)
y - 8 = 13 (x - 3)
y - 8 = 13x - 39
y = 13x - 39 +8
y = 13x -31
Hello,

you have calculated correctly the equation of the tangent of the curve.

Use

$\int \frac{dy}{dx} dx=y=\int(2x^2-5)dx = \frac23 x^3 - 5x + C$

Since (3, 8) lies on the curve the coordinates must satisfy the equation:

$8 = \frac23 (3)^3 - 5 \cdot 3 + C~\iff~C = 5$

Therefore the equation of the curve is:

$y = \frac23 x^3 - 5x + 5$