Answer:
Explanation:
To find the equation of the normal line to the surface z = 2x^4y^7 at the point (-1,1,2), we need to find the gradient of the surface at that point.
The gradient of a surface is a vector that points in the direction of the steepest increase in the surface, and its magnitude is the rate of change of the surface in that direction. To find the gradient, we take the partial derivatives of the surface with respect to each variable and form a vector:
∇f = ( ∂f/∂x, ∂f/∂y, ∂f/∂z )
For z = 2x^4y^7, we have:
∂f/∂x = 8x^3y^7
∂f/∂y = 28x^4y^6
∂f/∂z = 0
So, at the point (-1,1,2), the gradient is:
∇f = ( ∂f/∂x, ∂f/∂y, ∂f/∂z ) = ( 8(-1)^3(1)^7, 28(-1)^4(1)^6, 0 ) = (-8,28,0)
This means that the normal to the surface at the point (-1,1,2) is the vector (-8,28,0). To find the equation of the normal line, we can use the point-normal form of the equation of a line:
(x - x0)/a = (y - y0)/b = (z - z0)/c
where (x0, y0, z0) is the point on the line, and (a, b, c) is the direction vector of the line.
In this case, we have:
(x + 1)/(-8) = (y - 1)/28 = (z - 2)/0
Since the z-component of the direction vector is 0, we can drop the last term in the equation. Solving for x and y, we get:
x = -1 - (1/4)y
y = 1 + 28/8t
where t is a parameter that can take any value. So the equation of the normal line is:
x = -1 - (1/4)y
y = 1 + 28/8t
z = 2
or in parametric form:
r(t) = (-1 - (1/4)(1 + 28/8t))i + (1 + 28/8t)j + 2k