How do you find the equation of a line tangent to a graph z = sqrt (60 - x^2 - 2y^2) at the point (3, 5, 1)?

Oct 17, 2015

$3 x + 10 y + z - 60 = 0$

Explanation:

In the case of the function of two variables, we have tangent plane(s). So, if you mean tangent plane, the equation is:

$z - f \left({x}_{0} , {y}_{0}\right) = {f}_{x} \left({x}_{0} , {y}_{0}\right) \left(x - {x}_{0}\right) + {f}_{y} \left({x}_{0} , {y}_{0}\right) \left(y - {y}_{0}\right)$

${f}_{x} = \frac{\partial z}{\partial x} = \frac{1}{2 \sqrt{60 - {x}^{2} - 2 {y}^{2}}} \cdot \left(- 2 x\right) = - \frac{x}{\sqrt{60 - {x}^{2} - 2 {y}^{2}}}$

${f}_{y} = \frac{\partial z}{\partial y} = \frac{1}{2 \sqrt{60 - {x}^{2} - 2 {y}^{2}}} \cdot \left(- 4 y\right) = - \frac{2 y}{\sqrt{60 - {x}^{2} - 2 {y}^{2}}}$

$z - 1 = - \frac{3}{1} \left(x - 3\right) - \frac{10}{1} \left(y - 5\right)$

$z - 1 = - 3 x + 9 - 10 y + 50$

$3 x + 10 y + z - 60 = 0$

Every line that lies in the plane we've just found is tangent to a graph at a given point; there exists infinite number of such lines.