Using the gradient vector to find the tangent plane equation
Building the tangent plane equation from the gradient vector
We previously learned how to find the gradient vector at a specific point. We just use the formula
???\nabla{f(x,y)}=\left\langle\frac{\partial{f}}{\partial{x}},\frac{\partial{f}}{\partial{y}}\right\rangle???
where ???(x,y)??? is the point we’re interested in.
If the result of evaluating the gradient vector at the point ???(x,y)??? gives us
???\nabla{f(x,y)}=\left\langle a,b\right\rangle???
then ???a??? and ???b??? represent the slope of the original function in the ???x??? and ???y??? directions, respectively. Therefore, if we’re interested in finding the equation of the tangent plane at ???P(x_0,y_0)???, then we can plug the values of ???a??? and ???b???, along with the point ???P(x_0,y_0)???, into the formula for the equation of the tangent plane
???a(x-x_0)+b(y-y_0)-(z-z_0)=0???
where ???a??? and ???b??? come from ???\nabla{f(x,y)}=\left\langle{a},b\right\rangle???, ???x_0??? and ???y_0??? come from the given point ???P(x_0,y_0)???, and ???z_0??? is found by plugging ???P(x_0,y_0)??? into ???f(x,y)???.
Remember that the gradient vector and the equation of the tangent plane are not limited to two variable functions. We can modify the two variable formulas to accommodate more than two variables as needed.
If we have a function in three variables, the gradient vector is
???\nabla{f(x,y,z)}=\left\langle\frac{\partial{f}}{\partial{x}},\frac{\partial{f}}{\partial{y}},\frac{\partial{f}}{\partial{z}}\right\rangle???
where ???(x,y,z)??? is the point we’re interested in. If the result of evaluating the gradient vector at the point ???(x,y,z)??? gives us
???\nabla{f(x,y,z)}=\left\langle a,b,c\right\rangle???
then ???a???, ???b???, and ???c??? represent the slope of the original function in the ???x???, ???y???, and ???z??? directions, respectively. Therefore, if we’re interested in finding the equation of the tangent plane at ???P(x_0,y_0,z_0)???, then we can plug the values of ???a???, ???b???, and ???c???, along with the point ???P(x_0,y_0,z_0)???, into the formula for the equation of the tangent plane
???a(x-x_0)+b(y-y_0)-c(z-z_0)=0???
where ???a???, ???b???, and ???c??? come from ???\nabla{f(x,y,z)}=\left\langle{a},b,c\right\rangle???, and ???x_0???, ???y_0???, and ???z_0??? come from the given point ???P(x_0,y_0,z_0)???.
How to use the gradient vector for a function to find the tangent plane equation for the function at a particular point
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Find the gradient vector first, then use it to find the tangent plane equation
Example
Find the gradient vector of the function and use it to find the equation of the tangent plane at ???P(3,4)???.
???x^4-5x^3y-y^2+3y^4=6???
Rearrange the function.
???f(x,y)=x^4-5x^3y-y^2+3y^4-6???
Now we’ll start with the partial derivatives of the given function ???f???.
???\frac{\partial{f}}{\partial{x}}=4x^3-15x^2y???
and
???\frac{\partial{f}}{\partial{y}}=-5x^3-2y+12y^3???
To find the gradient vector at the point we’re interested in, we’ll plug the partial derivatives in to the formula for the gradient vector, and then evaluate at the point of interest.
???\nabla{f(x,y)}=\left\langle\frac{\partial{f}}{\partial{x}},\frac{\partial{f}}{\partial{y}}\right\rangle???
???\nabla{f(x,y)}=\left\langle4x^3-15x^2y,-5x^3-2y+12y^3\right\rangle???
Evaluating at ???P(3,4)???, we get
???\nabla{f(3,4)}=\left\langle4(3)^3-15(3)^2(4),-5(3)^3-2(4)+12(4)^3\right\rangle???
???\nabla{f(3,4)}=\left\langle-432,625\right\rangle???
Now we have everything we need to find the equation of the tangent plane.
???a(x-x_0)+b(y-y_0)-(z-z_0)=0???
???-432(x-3)+625(y-4)-(z-z_0)=0???
???z-z_0=-432x+1,296+625y-2,500???
???z-z_0=-432x+625y-1,204???
To find ???z_0???, we have to plug ???P(3,4)??? into ???f(x,y)???.
???f(3,4)=(3)^4-5(3)^3(4)-(4)^2+3(4)^4-6???
???f(3,4)=81-540-16+768-6???
???f(3,4)=287???
Plugging ???f(3,4)=287??? in for ???z_0??? gives
???z-287=-432x+625y-1,204???
???z=-432x+625y-917???