Write an equation of the tangent line `(5,4) at the point (x0,y0) := (5,4) to the level curve of f passing through (x0,y0).
F(x,y) = sqrt(x^2+(y^2/4)-6x+1)
Dont know how to do this, I may think I need to use the gradient vector, but dont know how.
Write an equation of the tangent line `(5,4) at the point (x0,y0) := (5,4) to the level curve of f passing through (x0,y0).
F(x,y) = sqrt(x^2+(y^2/4)-6x+1)
Dont know how to do this, I may think I need to use the gradient vector, but dont know how.
You use the words "gradient vector". Do you know what that is? Do you know that the gradient vector to z= sqrt(x^2+(y^2/4)-6x+1), at (x_0, y_0, z(x_0, y_0)) is? If you do know that the gradient vector, at any point, points in the direction of fastest increase of f and so is perpendicular to a curve on which there is no increase or decrease, the (level curve), you should see that you need to find a line through the given point. perpendicular to the gradient vector, that still lies in the tangent plane.
To be honest, I do not really know what gradient vector is. Nn my assignment that i handed in, the answer I got was completely wrong and I was told to take a look at gradient in my Math book. The sqrt part really confuses me, thats what ultimately stops me from getting this problem solved.
First, at x= 5, y= 4, f(5, 4)= 25+ 4- 30+ 1= 0 so this level curve is sqrt(x^2+ (y^2/4)- 6x+ 1)= 0 which is the same as x^2+ (y^2/4)- 6x+ 1= 0. That is why Idea suggested squaring the given function.
Since this is a level curve, rather than the surface itself, and lies entirely in the xy-plane, no, you do not need the "gradient vector". We do need to calculate the derivative of this "implicit" function: 2x+ (y/2)y'- 6= 0 so that y'= (2/y)(6- 2x). At x= 5, y= 4, y'= (2/4)(6- 10)= -2. Do you know how to write the tangent line to that curve at (5, 4)?