14.6 - Tangent Planes Revisited

Recall that $\nabla f$ is orthogonal to the level curves/surfaces.

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Example

Example

$w=f(x,y,z)$

• This function has the level surfaces $f(x,y,z)=K$ where K is constant
• Let C be any curve on the level surface
• Suppose C is parameterized by some vector function $\vec{r}(t)=⟨x(t),y(t),z(t)⟩$

If this is the case, then we can say something like:

$f(x(t),y(t),z(t))=K$

We can use the chain rule here to find the $\frac{\partial }{\partial t}$ of both sides

Applying this onto our tree diagram method:

graph TD
    f --> x
    f --> y
	f --> z
    x --> tx
    z --> tz
    y --> ty

Setting up the equation:

$f_x{x}_t+f_y{y}_t+f_z{z}_t=0$

Putting this in unit vector form:

$⟨f_x,f_y,f_z⟩\cdot ⟨x^{\prime }(t),y^{\prime }(t),z^{\prime }(t)⟩=0$

Which finally yields the key equation:

Important Information

Key Formula

$\nabla f\cdot {\vec{r}}^{\prime }(t)=0$

Where the key formula is $\sqrt{IF}$

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What This Tells Us

This key formula tells us the the gradient vector(at any point) is orthogonal to the tangent vector (at the same point) to any curve on the surface passing through that point.

Result of this action: $\nabla f$ is a normal vector for the tangent plane.

For the surface $f(x,y,z)=K$, the tangent plane at the point $(x_0,y_0,z_0)$ is:

$f_x(x_0,y_0,z_0)(x-x_0)+f_y(x_0,y_0,z_0)(y-y_0)+f_z(x_0,y_0,z_0)(z-z_0)=0$

Which then yields:

$\nabla f\cdot \nabla r=0$

$\nabla r=⟨\nabla x,\nabla y,\nabla z⟩=⟨x-x_0,y-y_0,z-z_0⟩$

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Tangent Line to Curve of Intersection of Surfaces

Find the tangent line to curve of intersection of:

$z=x^2+y^{2}4x^2+y^2+z^2=9at(-1,1,2)$