Question

math application of integrals question ​

Answer 1

Question :

Using integration, find the area of the triangle formed by positive x- axis and tangent and normal to the circle

${x}^{2} + {y }^{2} = 4 \: \: at \: (1, \sqrt{3} )$

Solution:

The given circle is

${x}^{2} + {y }^{2} = 4$

Differentiate w.r.t.x

$2x + 2y \frac{dy}{dx} = 0 \\ \\ = > \frac{dy}{dx} = \frac{ - x}{y}$

Slope of tangent at P (1,√3) = -1/√3

Slope of normal at P(1,√3)= √3

Therefore,

The equation of the tangent at P is :

$y - \sqrt{3} = - \frac{1 }{ \sqrt{3} } (x - y) \\ \\ = > y = - \frac{ 1}{ \sqrt{3} } x + \sqrt{3} + \frac{1}{ \sqrt{3} }$

This meets x-axis, where....

$0 = - \frac{1}{3} x + \sqrt{3} + \frac{1}{ \sqrt{3} } \\ \\ = > 0 = - x + 3 + 1 \\ \\ = > x = 4$

Thus T is (4,0)

The equation of the normal at P is :

$y - \sqrt{3} = \sqrt{3} (x - 1) \\ \\ y = \sqrt{3x}$

This meets x-axis, y=0

Thus O is (0,0)

Now ar(△OPT)

= ar(OPL)+ar(PLT)

$=\int\limits_{0}^{1} \sqrt{3x} \: dx + \int\limits_{1}^{4}( - \frac{1}{ \sqrt{3} } x + \sqrt{3} + \frac{1}{ \sqrt{3} } )dx \\ \\ = \sqrt{3} \int\limits_{0}^{1}x \: dx - \frac{1}{ \sqrt{3} } \int\limits_{1}^{4}x \: dx +( { \sqrt{3} + \frac{1}{ \sqrt{3} } } ) \int\limits_{1}^{4}(1)dx \\ \\ = \sqrt{3} \: \: [ \frac{ {x}^{2} }{2} ]limit0→1 \: \: - \frac{1}{ \sqrt{3} } [ \frac{ {x}^{2} }{2} ]limit1→4 \: \: + \frac{3 + 1}{ \sqrt{3} }[ {x}^{2} ]limit1→4 \\ \\ = \sqrt{3} ( \frac{1}{2} - 0) - \frac{1}{ \sqrt{3} } (8 - \frac{1}{2} ) + \frac{4}{ \sqrt{3} } (4 - 1) \\ \\ = \frac{ { \sqrt{3} } }{2} - \frac{8}{ \sqrt{3} } + \frac{1}{ 2 \sqrt{3} } + 4 \sqrt{3} \\ \\ = \frac{9}{2} \sqrt{3} + \frac{ - 16 + 1}{2 \sqrt{3} } \\ \\ = \frac{9}{2} \sqrt{3} - \frac{15}{2 \sqrt{3} } \\ \\ = \frac{9}{2} \sqrt{3} - \frac{5}{2} \sqrt{3} \\ \\ = ( \frac{9}{2} - \frac{5}{2} ) \sqrt{3} \\ \\ = 2 \sqrt{3} sq \: units.$