Question

Find parametric equations for the normal line to the surface z=7x 2 −4y 2 at the point (2,1,25)

A x=2t+28,y=t+14,z=24t−1
B x=2t+14,y=t−4,z=−t+24
C x=28t+2,y=−8t+1,z=−t+24
D x=14t+2,y=−4t+1,z=−t+10
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Answer 1

To find the parametric equations for the normal line, we calculate the gradient of the surface to get the direction of the normal vector at the point (2,1,25). The equations of the normal line turn out to be x = 28t + 2, y = -8t + 1, z = -1t + 24.

Step-by-step explanation:

The question asks to find parametric equations for the normal line to the surface z=7x2 - 4y2 at the point (2,1,25). First, we need to calculate the gradient of the surface, which gives us the direction of the normal vector at that point. The gradient is found by taking the partial derivatives of the surface equation.

The partial derivative with respect to x is ∂z/∂x = 14x, and at x=2, it is 28. Similarly, the partial derivative with respect to y is ∂z/∂y = -8y, which at y=1 is -8. So the gradient at the point (2,1,25) is (28, -8, 1).

The normal line's parametric equations will have the direction of this gradient, so the components of the normal vector will act as the coefficients of 't' in the equations. Hence,

• x = 28t + 2
• y = -8t + 1
• z = -1t + 24

These parametric equations represent the normal line to the given surface at the specified point.