Find an equation of the plane that is tangent to the given surface at the specified pointz=3y2−2x2+x,&nbsp;(2,−1,−3)

Question

Find an equation of the plane that is tangent to the given surface at the specified pointz=3y2−2x2+x,&amp;nbsp;(2,−1,−3)

Aamina Herring · Accepted Answer

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)[x−2]+fy(2,−1)[y+1] Plug in respective values z+3=−7x+14−6y−6 Simplify −7x−6y+5=z

Jeffrey Jordon · Accepted Answer

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 &amp;lt;-7,-6,-1&amp;gt;.   The equation of the tangent plane is given by −7(x−2)−6(y+1)−(z+3)=0  which simplifies to z=−7x−6y+5

RizerMix · Accepted Answer

To solve the equation z=3y2−2x2+x for the point (2,−1,−3), we can substitute x=2 and y=−1 into the equation and simplify:z=3y2−2x2+x=3(−1)2−2(2)2+2=−3−8+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=3y2−2x2+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:−7x−6y+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 −7x−6y+5=z.

Vasquez · Accepted Answer

Starting with the given equation:z=3y2−2x2+xWe can substitute the values for x and y from the given point (2,−1,−3) to get:z=3(−1)2−2(2)2+2=−7So the point (2,−1,−3) satisfies the equation z=−7.To write the equation in the form z=−7x−6y+5, we can solve for x in terms of y and z.Starting with the given equation:z=3y2−2x2+xRearranging and completing the square for x:2x2−x=z−3y22(x−14)2−18=z−3y22(x−14)2=z−3y2+18x−14=±z−3y2+182x=14±z−3y2+182Substituting this into the equation z=−7:−7=14±−7−3y2+182−7−14=±−7−3y2+182−294=±−28−12y2+116−294=±−12y2−2716Squaring both sides and simplifying:84116=−12y2−271612y2=−868y2=−2173This is not a real solution for y, so the point (2,−1,−3) does not satisfy the equation z=−7x−6y+5.Therefore, there is no solution to the given equation that satisfies the point (2,−1,−3) and is in the form z=−7x−6y+5.

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

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