Tangent plane to $x^2+y^2+z^2=50$ at $(3,4,5)$

Question

Prove that the tangent plane to $x^2+y^2+z^2=50$ at $(3,4,5)$ is $3x+4y+5z=50$

My workings are shown below but get the answer wrong completly, have I made a simple mistake is my method flawed.

So if the surface is said to be $f(x,y,z)$ then $$\frac{\partial f}{\partial x}=2x\qquad \frac{\partial f}{\partial y}=2y$$

Therefore $f_x(3,4)=6$ and $f_y(3,4)=8$

Now using the $$z=f(a,b)+(x-a)f_x(a,b)+(y-b)f_y(a,b)$$ I get a plane with the equation $$z=5+(x-3)(6)+(y-4)(8)$$ $$z=6x+8y-45$$

Any help on how to get the correct answer would be apprciated.

Answer 1

A few comments:

• The formula you are referring to is for surfaces which are the graphs of a function $f(x,y)$, not $f(x,y,z)$
• The given surface is a sphere of radius $\sqrt{50}$ centered at the origin. A sphere cannot be represented as a function $f(x,y)$

However, hope is not lost. The given point $(3,4,5)$ lies on the upper hemisphere, and a hemisphere is the graph of a function $z=f(x,y)$. What function is it? Well in this case we have $$z^{2}=50-x^{2}-y^{2}$$ so the top hemisphere is given by $$z=\sqrt{50-x^{2}-y^{2}}=(50-x^{2}-y^{2})^{1/2}$$ This is our $f(x,y)$. I'll point out that the lower hemisphere is given by $-f(x,y)$. Now, with this $f$, we have $$\frac{\partial f}{\partial x}=\frac{1}{2}(50-x^{2}-y^{2})^{-1/2}\cdot(-2x)$$ and $$\frac{\partial f}{\partial y}=\frac{1}{2}(50-x^{2}-y^{2})^{-1/2}\cdot(-2y)$$

Plugging in $x=3,y=4$ we have $$f_{x}(3,4)=-\frac{3}{5}$$ and $$f_{y}(3,4)=-\frac{4}{5}$$ Since $f(3,4)=5$, the tangent plane has equation $$z=5-\frac{3}{5}(x-3)-\frac{4}{5}(y-4)$$ which, after some algebra, becomes $$3x+4y+5z=50$$

Answer 2

In order to calculate the tangent plane at $P = (3,4,5)$, we can consider the parametrization:

\begin{align*} \varphi(x,y) = \left(x,y,\sqrt{50 - x^{2} - y^{2}}\right) \end{align*}

Consequently, the tangent plane $\pi$ at the given point can be described by

\begin{align*} (x,y,z) = P + \alpha\varphi_{x}(3,4) + \beta\varphi_{y}(3,4) \end{align*}

where $(\alpha,\beta)\in\mathbb{R}^{2}$.

Answer 3

$f(x,y,z)=x^2+y^2+z^2-50=0;$

Normal vector to surface $f(x,y,z)=0$ is given by

$\vec n= \nabla f(x,y,z)=(f_x,f_y,f_z)=(2x,2y,2z)$

Equation.of tangent plane:

$\vec n \cdot (\vec r -\vec r_0)=0$;

With $\vec n= 2(3,4,5)$ and $\vec r_0=(3,4,5)$:

$2(3,4,5)\cdot ((x,y,z)-(3,4,5))=0;$

$3x+4y+5z -9-16-25=0;$

$3x+4y+5z =50.$

Answer 4

Why not forget Calculus and look geometrically?

Having observed that the point $P=(3,4,5)$ is indeed on the sphere, one sees that the segment $\overline{\Bbb OP}$ is a radius, where $\Bbb O$ is the origin, center of the sphere. Now certainly any radius is perpendicular to the tangent plane at the point in question, so that the vector $(3,4,5)$ is orthogonal to the plane, and thus the equation must be $3X+4Y+5Z=c$, for an appropriate $c$, which you find to be $c=50$, in order to guarantee that $P$ be on the plane.