{"id":5595,"date":"2021-09-22T16:46:59","date_gmt":"2021-09-22T11:16:59","guid":{"rendered":"https:\/\/mathemerize.com\/?p=5595"},"modified":"2021-11-13T20:07:59","modified_gmt":"2021-11-13T14:37:59","slug":"angle-of-intersection-of-two-curves","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/angle-of-intersection-of-two-curves\/","title":{"rendered":"Angle of Intersection of Two Curves"},"content":{"rendered":"
Here you will learn angle of intersection of two curves formula with examples.<\/p>\n
Let’s begin –<\/p>\n
The angle of intersection of two curves is defined to be the angle between the tangents to the two curves at their point of intersection.<\/p>\n
Let \\(C_1\\) and \\(C_2\\) be two curves having equations y = f(x) and y = g(x) respectively.<\/p>\n
and \\(m_1\\) = slope of tangent to y = f(x) at P = \\(({dy\\over dx})_{C_1}\\)<\/p>\n
and \\(m_2\\) = slope of the tangent to y = g(x) at P = \\(({dy\\over dx})_{C_2}\\)<\/p>\n
\nAngle between the curve is \\(tan \\phi\\) = \\(m_1 – m_2\\over 1 + m_1 m_2\\)<\/p>\n<\/blockquote>\n
Orthogonal Curves <\/h3>\n
If the angle of two curves is at right angle, the two curves are equal to intersect orthogonally and the curves are called orthogonal curves.<\/p>\n
\nIf the curves are orthogonal then \\(\\phi\\) = \\(\\pi\\over 2\\)<\/p>\n
\\(\\therefore\\) \\(m_1 m_2\\) = -1<\/p>\n<\/blockquote>\n
Note<\/strong> : Two curves \\(ax^2 + by^2\\) = 1 and \\(a’x^2 + b’y^2\\) = 1 will intersect orthogonally, if<\/p>\n
\n\\(1\\over a\\) – \\(1\\over b\\) = \\(1\\over a’\\) – \\(1\\over b’\\)<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : find the angle between the curves xy = 6 and \\(x^2 y\\) =12.<\/p>\n
Solution<\/span><\/strong> : The equation of the two curves are<\/p>\n
xy = 6 …….(i)<\/p>\n
and, \\(x^2 y\\) = 12 …………(ii)<\/p>\n
from (i) , we obtain y = \\(6\\over x\\). Putting this value of y in (ii), we obtain<\/p>\n
\\(x^2\\) \\((6\\over x)\\) = 12 \\(\\implies\\) 6x = 12 <\/p>\n
\\(\\implies\\) x = 2<\/p>\n
Putting x = 2 in (i) or (ii), we get y = 3. <\/p>\n
Thus, the two curves intersect at P(2, 3).<\/p>\n
Differentiating (i) with respect to x, we get<\/p>\n
x\\(dy\\over dx\\) + y = 0 \\(\\implies\\) \\(dy\\over dx\\) = \\(-y\\over x\\)<\/p>\n
\\(\\implies\\) \\(m_1\\) = \\(({dy\\over dx})_{(2, 3)}\\) = \\(-3\\over 2\\)<\/p>\n
Differentiating (ii) with respect to x, we get<\/p>\n
\\(x^2\\) \\(dy\\over dx\\) + 2xy = 0 \\(\\implies\\) \\(dy\\over dx\\) = \\(-2y\\over x\\)<\/p>\n
\\(\\implies\\) \\(m_2\\) = \\(({dy\\over dx})_{(2, 3)}\\) = -3<\/p>\n
Let \\(\\theta\\) be the angle, then<\/p>\n
\\(tan \\theta\\) = \\(m_1 – m_2\\over 1 + m_1 m_2\\) = \\(3\\over 11\\)<\/p>\n
\\(\\theta\\) = \\(tan^{-1} (3\/11)\\)<\/p>\n
\nRelated Questions<\/h3>\n
The angle of intersection between the curve \\(x^2\\) = 32y and \\(y^2\\) = 4x at point (16, 8) is<\/a><\/p>\n
Check the orthogonality of the curves \\(y^2\\) = x and \\(x^2\\) = y.<\/a><\/p>\n\n\n