Question
Download Solution PDFEliminating a and b from the (x - a)2 + (y - b)2 + z2 = c2 the partial differential equation is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFExplanation:
(x - a)2 + (y – b)2 + z2 = c2 ---(i)
Differentiating (i) w.r.t ‘x’
\(2\left( {x - a} \right) + 0 + 2z\frac{{\partial z}}{{\partial x}} = 0\)
\(2 \left( {x - a} \right) = - 2z\frac{{\partial z}}{{\partial x}}\)
\(\left( {x - a} \right) = - z\frac{{\partial z}}{{\partial x}} = \phi \)
Now,
Differentiating (i) w.r.t. ‘y’
\(2 \left( {y - b} \right) +2z\frac{{\partial y}}{{\partial z}} = 0\)
\(2\left( {y - b} \right) = - 2z\frac{{\partial y}}{{\partial z}}\)
\(\left( {y - b} \right) = - z\frac{{\partial y}}{{\partial z}}\)
\({\left( { - z\frac{{\partial z}}{{\partial x}}} \right)^2} + {\left( { - z\frac{{\partial y}}{{\partial z}}} \right)^2} + {z^2} = {c^2}\)
(-zp)2 + (-zq)2 + z2 = c2
z2 (p2 + q2 + 1) = c2
∴ Option (2) is correct.
Last updated on Jun 19, 2025
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