Respuesta :
The tangent planes to the surface are described by the equation "6x + 12y + 48z - 210 = 0" (3, 4, 3).
In this instance, the equation yields the parametrized surface "r(t) = (t², [tex]t^3[/tex], [tex]t^4[/tex]). On the surface, the issue that interests us is (3, 4, 3).
We may use the partial derivative concept to obtain the second derivative of the surface at points (3, 4, 3). Regardless of whether a change in the parameterization approaches zero, the time derivative of the parameter value of a variable serves as the upper estimate of the differentiating quotient.
[(r(3 + h) - r(3)) / h] as h approaches 0
For the x-coordinate of the surface, this limit is given by
[((3 + h)² - 3²) / h] as h approaches 0
which simplifies to
[(9 + 6h + h² - 9) / h] as h approaches 0
which simplifies to
[6h + h²] as h approaches 0
Since h is approaching 0, h2 is approaching it faster than h. As h approaches 0, the limit changes from 0 to 6, and the number of h2 decreases relative to the value of 6h.
Thus, the partial derivative of the x-coordinate of the surface concerning the time at the coordinates (3, 4, 3) is 6.
As a result, the vector is the normal vector to the ground at the location (3, 4, 3). (6, 12, 48). 6x + 12y + 48z + d = 0 is the equation of the tangent planes to the ground at this position.
Learn more about the tangent plane at
https://brainly.com/question/29589716
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