Calculus: Tangents & Normals
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parametric differen.
equation of a tangent
equation of a normal
rate of change prob.1
rate of change prob.2
the Chain Rule
Chain Rule probs. #1
Chain Rule probs. #2
Chain Rule probs. #3
the Product Rule
parametric eqs.prob#1
parametric eqs.prob#2
intro. to integration
integration by parts 1
integration by parts 2
area under a curve
volumes of revolution
area between curves
Binomial Theorem
Bin. Theorem problems
Trig. Identities
Half Angle Formula
Double Angle Formula
Vector problems
Vector problems in 3D
MORE . . .


derivative formula
tangents & normals
inverse functions
MORE . . .


Edxl C1 Pure specimen
Edxl C1 Pure answers
Edxl C2 Pure specimen
Edxl C2 Pure answers
Edxl C3 Pure specimen
Edxl C3 Pure answers
Edxl C4 Pure specimen
Edxl C4 Pure answers
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the derivative formula
tangents & normals
maxima & minima
chain rule
diffn.trigonometric fns.
product rule
quotient rule
parametric equations
implicit equations
differential equations
integration formula
int.'by substitution'
int.'by parts'
algebraic fractions
definite integrals
areas under curves
volumes of revolution
Trapezium Rule
integ. diff. equations
sine, cosine, tangent
sec, cosec, cotan
Sine & Cosine Rules
Pythagorean ID's
compound angles
Sigma Notation
arithmetic progression
geometic progression
line between points
more straight lines
parametric equations
circles & ellipses
MORE . . .

[ tangents ][ normals ][ equation tan. ][ equation normal ]






The gradient of the tangent to the curve y = f(x) at the point (x1, y1) on the curve is given by:

the value of dy/dx, when x = x1 and y = y1


tangent and normal


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Two lines of gradients m 1, m 2 respectively are perpendicular to eachother if the product,

m 1x m 2 = -1


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Equation of a tangent

The equation of a tangent is found using the equation for a straight line of gradient m, passing through the point (x1, y1)

y - y1= m(x - x1)

To obtain the equation we substitute in the values for x1 and y1 and m (dy/dx) and rearrange to make y the subject.


Find the equation of the tangent to the curve y = 2x2 at the point (1,2).

tangent to curve problem #1


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Equation of a normal

The equation of a normal is found in the same way as the tangent. The gradient(m 2 )of the normal is calculated from;

m 1x m 2 = -1 (where m 1 is the gradient of the tangent)


m 2 = - 1/( m 1)



Find the equation of the normal to the curve:

  y = x2 + 4x + 3,   at the point (-1,0).

equation of a normal to a curve #1






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