Find an equation of the tangent plane to the given surface at the specified point. z=3y^2-2x^2+x,\ (2,-1,-3)

avissidep

avissidep

Answered question

2021-05-27

Find an equation of the plane that is tangent to the given surface at the specified pointz=3y22x2+x, (2,1,3)

Answer & Explanation

Aamina Herring

Aamina Herring

Skilled2021-05-28Added 85 answers

Solve tfor the partial derivative of f with respect to x
fx(x,y)=4x+1
fy(x,y)=6y
At the point (2,1) evaluate
fx=(2,1)=7
At the point (2,-1) evaluate
fy(2,1)=6
Frame the equation of the tangent plane
z+3=fx(2,1)[x2]+fy(2,1)[y+1]
Plug in respective values
z+3=7x+146y6
Simplify
7x6y+5=z

Jeffrey Jordon

Jeffrey Jordon

Expert2021-10-14Added 2605 answers

You can use the partial derivatives to find a normal vector to the surface. At the point (2,-1,-3) you have fx(2,1)=7 and fy(2,1)=6 so that one normal vector is <-7,-6,-1>. 

The equation of the tangent plane is given by 7(x2)6(y+1)(z+3)=0

which simplifies to z=7x6y+5

RizerMix

RizerMix

Expert2023-04-30Added 656 answers

To solve the equation z=3y22x2+x for the point (2,1,3), we can substitute x=2 and y=1 into the equation and simplify:
z=3y22x2+x
=3(1)22(2)2+2
=38+2
=9.
Therefore, the solution is z=9 when x=2 and y=1. To write the answer in the requested format, we can rewrite the equation as:
z=3y22x2+x
=2x2+x+3y2
=2(2)2+2+3(1)2
=8+2+3
=3.
So, z=3 when x=2 and y=1. To convert this into the requested format, we can write:
7x6y+5=7(2)6(1)+5
=14+6+5
=3,
which confirms that z=3 when x=2 and y=1. Therefore, the final solution is 7x6y+5=z.
Vasquez

Vasquez

Expert2023-04-30Added 669 answers

Starting with the given equation:
z=3y22x2+x
We can substitute the values for x and y from the given point (2,1,3) to get:
z=3(1)22(2)2+2=7
So the point (2,1,3) satisfies the equation z=7.
To write the equation in the form z=7x6y+5, we can solve for x in terms of y and z.
Starting with the given equation:
z=3y22x2+x
Rearranging and completing the square for x:
2x2x=z3y2
2(x14)218=z3y2
2(x14)2=z3y2+18
x14=±z3y2+182
x=14±z3y2+182
Substituting this into the equation z=7:
7=14±73y2+182
714=±73y2+182
294=±2812y2+116
294=±12y22716
Squaring both sides and simplifying:
84116=12y22716
12y2=868
y2=2173
This is not a real solution for y, so the point (2,1,3) does not satisfy the equation z=7x6y+5.
Therefore, there is no solution to the given equation that satisfies the point (2,1,3) and is in the form z=7x6y+5.

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