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)?

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Answer 1 · 2015-10-17T17:56:43

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

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(x_0,y_0)=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)#

#f_x=(delz)/(delx)=1/(2sqrt(60-x^2-2y^2))*(-2x) = -x/sqrt(60-x^2-2y^2)#

#f_y=(delz)/(dely)=1/(2sqrt(60-x^2-2y^2))*(-4y) = -(2y)/sqrt(60-x^2-2y^2)#

#z-1=-3/1(x-3)-10/1(y-5)#

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

#3x+10y+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.