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)?

Question

Cameron Alexander · Accepted Answer

#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.

William Abdalla · Answer

undefined To find the equation of a line tangent to a graph at a given point, we need to find the gradient vector at that point.

First, we find the partial derivatives of the given function with respect to x and y:

∂z/∂x = (-x) / sqrt(60 - x^2 - 2y^2)
∂z/∂y = (-2y) / sqrt(60 - x^2 - 2y^2)

Next, we evaluate these partial derivatives at the point (3, 5, 1):

∂z/∂x = (-3) / sqrt(60 - 3^2 - 2(5^2))
∂z/∂y = (-2 * 5) / sqrt(60 - 3^2 - 2(5^2))

Simplifying further:

∂z/∂x = -3 / sqrt(20)
∂z/∂y = -10 / sqrt(20)

The gradient vector at the point (3, 5, 1) is given by:

∇z = (∂z/∂x, ∂z/∂y) = (-3 / sqrt(20), -10 / sqrt(20))

Finally, we can write the equation of the tangent line using the point-normal form:

(x - 3) / (-3 / sqrt(20)) = (y - 5) / (-10 / sqrt(20)) = (z - 1) / 1