Application of Differentiation
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Find equation of tangent and normals to the curve.
Tangents and normals have perpendicular gradients
(m1 x m2 = -1)
Tangents and normals are both straight lines
meaning y – y1 = m ( x – x1)
Example:
The curve y = x^3 – x^2 passes through P(2, 4), find the equation of the tangent at this point.
y = x^3 – x^2 dy/dx = 3x^2 – 2x
At P, x = 2 therefore the gradient = 3(2)^2 – 2(2) = 8
m=8 y=4 x=2
y – 4 = 8(x -2)
y = 8x – 12
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Finding a stationary point
Increasing: dy/dx > 0
Decreasing: dy/dx < 0
Turning Point: dy/dx = 0
To identify if maximum or minimum use
d^2y/dx^2 or consider dy/dx either side of the
stationary point
dy/dx = 0 d^2y/dx^2 = 0
---------------------------------------------------------------------------------------------------------
Find equation of tangent and normals to the curve.
Tangents and normals have perpendicular gradients
(m1 x m2 = -1)
Tangents and normals are both straight lines
meaning y – y1 = m ( x – x1)
Example:
The curve y = x^3 – x^2 passes through P(2, 4), find the equation of the tangent at this point.
y = x^3 – x^2 dy/dx = 3x^2 – 2x
At P, x = 2 therefore the gradient = 3(2)^2 – 2(2) = 8
m=8 y=4 x=2
y – 4 = 8(x -2)
y = 8x – 12
---------------------------------------------------------------------------------------------------------
Finding a stationary point
Increasing: dy/dx > 0
Decreasing: dy/dx < 0
Turning Point: dy/dx = 0
To identify if maximum or minimum use
d^2y/dx^2 or consider dy/dx either side of the
stationary point
dy/dx = 0 d^2y/dx^2 = 0
