Question

\({x^{\rm{2}}} + {y^{\rm{2}}} = {r^{\rm{2}}}\), \(ax + by = {\rm{0}}\).

Short answer

The family of curves \({x^{\rm{2}}} + {y^{\rm{2}}} = {r^{\rm{2}}}\) and \(ax + by = {\rm{0}}\)are orthogonal trajectories of each other and their sketch is,

Step-by-step solution

Given Information

The given family of curves are \({x^{\rm{2}}} + {y^{\rm{2}}} = {r^{\rm{2}}}\)and \(ax + by = {\rm{0}}\).

Definition of Derivative

The derivative of a real-valued function measures the sensitivity of the function's value (output value) to changes in its argument in mathematics (input value).

Slopes of family of curves

Differentiate and find the slope of curves.

Differentiate the curve\({x^{\rm{2}}} + {y^{\rm{2}}} = {r^{\rm{2}}}\) with respect to \(x\).

\(\begin{array}{c}{x^{\rm{2}}} + {y^{\rm{2}}} = {r^{\rm{2}}}\\{\rm{2}}x + {\rm{2}}y{{y'}_{\rm{1}}} = {\rm{0}}\\{{y'}_{\rm{1}}} = - \frac{x}{y}\end{array}\)

Find the ratio of constants in curve \(ax + by = {\rm{0}}\) and get \( - \frac{a}{b} = \frac{y}{x}\).

Differentiate the curve \(ax + by = {\rm{0}}\) with respect to \(x\).

\(\begin{array}{c}ax + by = {\rm{0}}\\a + b{{y'}_{\rm{2}}} = {\rm{0}}\\{{y'}_{\rm{2}}} = - \frac{a}{b}\\{{y'}_{\rm{2}}} = \frac{y}{x}\end{array}\)

Orthogonality of family of curves

The product of slopes of two orthogonal slopes is always equals to \( - {\rm{1}}\).

Find the product of slopes of the curves.

\(\begin{array}{c}{{y'}_{\rm{1}}}{{y'}_{\rm{2}}} = \left( { - \frac{x}{y}} \right)\left( {\frac{y}{x}} \right)\\ = - {\rm{1}}\end{array}\)

So, the two curves are orthogonal to each other

Sketch of curves

Sketch the graph of the given family of curves on the same axes and show their orthogonal trajectories

Therefore, the family of curves \({x^{\rm{2}}} + {y^{\rm{2}}} = {r^{\rm{2}}}\) and \(ax + by = {\rm{0}}\) are orthogonal trajectories of each other and their sketch is,