Tangents and Normals by Differentiation

Any point P(x, y) on a curve has a tangent line that denotes the gradient at point P. By differentiating the equation of the curve, and substituting (x, y) to the resulting equation, you would get the value of the gradient at point P.

The gradient of the tangent to a function (or curve) is denoted by \cfrac{dy}{dx}.

 

Examples:

1. Find the gradient of the tangent to the parabola y=x^2+1 at the point (1, 2).

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\cfrac { dy }{ dx } =2x\\ \\ At\quad (1,2),\quad \cfrac { dy }{ dx } =2(1)=2

So the gradient of the tangent at (1, 2) is 2.

 

 

 

2. Find the values of x for which the gradient of the tangent to the curve y=2{x}^{3}-6{x}^{2}+1 is equal to 18.

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\cfrac { dy }{ dx } =6{ x }^{ 2 }-12x\\ \\ Substitute:\quad \cfrac { dy }{ dx } =18\\ \\ 18=6{ x }^{ 2 }-12x\\ 6{ x }^{ 2 }-12x-18=0\\ { x }^{ 2 }-2x-3=0\\ (x-3)(x+1)=0\\ \\ Hence,\quad x=3,\quad or\quad x=-1

 

 

Explore tangents and normals with the Geogebra app below:

The normal is a straight line perpendicular to the tangent at the same point of contact with the curve.

If lines with gradients {m}_{1} and {m}_{2} are perpendicular, then {m}_{1}{m}_{2}=-1

 

Examples:

1. Find the gradient of the normal to the curve y=2{x}^{2}-3x+5 at the point where x = 4.

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\cfrac { dy }{ dx } \quad is\quad the\quad gradient\quad of\quad the\quad tangent.\\ \\ \cfrac { dy }{ dx } =4x-3\\ \\ When\quad x=4\\ \cfrac { dy }{ dx } =4(4)-3\\ \cfrac { dy }{ dx } =13\\ \\ So\quad { m }_{ 1 }=13\\ \\ The\quad normal\quad is\quad perpendicular\quad to\quad the\quad tangent.\\ \\ So\quad { m }_{ 1 }{ m }_{ 2 }=-1\\ \\ 13\quad \times \quad { m }_{ 2 }=-1\\ \\ { m }_{ 2 }=-\cfrac { 1 }{ 13 } \\ \\

So the gradient of the normal is -\cfrac { 1 }{ 13 }

 

 

2. Find the equation of the normal to the curve {x}^{3}+3{x}^{2}-2x-1 at the point (-1, 3).

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\cfrac { dy }{ dx } =3{ x }^{ 2 }+6x-2\\ \\ When\quad x=-1\\ \cfrac { dy }{ dx } =3({ -1 }^{ 2 })+6(-1)-2\\ \cfrac { dy }{ dx } =-5\\ \\ { m }_{ 1 }=-5\\ \\ So,\quad { m }_{ 1 }{ m }_{ 2 }=-1\\ \\ -5\quad \times \quad { m }_{ 2 }\quad =\quad -1\\ { m }_{ 2 }\quad =\quad \cfrac { 1 }{ 5 } \\ \\ So\quad the\quad gradient\quad of\quad the\quad normal\quad is\quad \cfrac { 1 }{ 5 } .\\ \\ Equation\quad of\quad the\quad normal:\\ y-{ y }_{ 1 }=m(x-{ x }_{ 1 })\\ y-3=\cfrac { 1 }{ 5 } (x-(-1))\\ 5y-15=x+1\\ x-5y+16=0