Find the (linear) equation of the tangent plane to the surface z=5y2−3x2+x at the point (2,−1,−5). Your answer should be in the form of an equation, e.g., something like ????x+????y+cz=???? or ????(x−x0)+????(y−y0)+c(z−z0)=0 would work.
Accepted Solution
A:
Answer:The equation of the tangent plane to the surface is given by [tex]6x+10y+z=-3[/tex]Step-by-step explanation:We can find the normal of the surface using the gradient over f(x,y,z), where the function is[tex]f(x,y,z)=-3x^2+x+5y^2-z=0[/tex]And the gradient is[tex]\nabla f(x,y,z) =<-3x+1, 10y, -1>[/tex]Then the normal at the point (2,-1, -5) is[tex]\vec n =\nabla f(2,-1,-5)\\\vec n = <-3(2), 10(-1), -1>\\\vec n = <-6,-10,-1>[/tex]Then the equation of the tangent plane to the surface is given by[tex]\vec n \cdot (P-P_0)=\vec0[/tex]Replacing the given point and the normal we get[tex]<-6,-10,-1>\cdot <x-2, y+1, z+5>=0\\-6(x-2)-10(y+1)-(z+5)=0[/tex]We can simplify a bit to get into standard form[tex]-6x+12-10y-10-z-5=0\\-6x-10y-z-3=0\\-6x-10y-z=3\\\\6x+10y+z=-3[/tex]