**–**$x^2 + 2y^2 + 3xz = 1-$**, at the point**$(1, 2, \dfrac{1}{3})$**–**$y^2 – x^2 = 3$**, at the point**(1,2,8)

This problem aims to find the 2D planes that are **tangent** to the given **surfaces**. To better understand the problem, you must be familiar with **tangents**, **normal** **lines**, and **linear approximation** techniques.

Now, **tangent** **planes** lying on a surface are **planes** that just **brush** a surface at some particular **point** and are also **parallel** to the surface at that point. One thing to note here is the **point** which lies on the **plane**. Let assume $(x_0, y_0, z_0)$ to be any point on the surface $z = f(x, y)$. If the **tangent** **lines** at $(x_0, y_0, z_0)$ to all **curves** on the **surface** departing through $(x_0, y_0, z_0)$ lie on a shared plane, that **plane** is known as a **tangent plane** to $z = f(x, y)$ at$(x_0, y_0, z_0)$.

## Expert Answer

The **formula** to find the **tangent** **plane** on a given smooth **curved** **surface** is:

\[\nabla f(x_0). (x -x_0)=0 \]

**Part a:**

\[f(x,y,z)=x^2 + 2y^2 + 3xz, x_0 = (1, 2, \dfrac{1}{3})\]

**Given** $f(x_0)=k$:

\[f(x_0)=1^2 + 2(2)^2 + 3(\dfrac{1}{3}) = 10\]

\[k=10\]

Now **calculating** $\nabla f(x)$:

\[\nabla f(x) = (\dfrac{d}{dx} (x^2 + 2y^2 + 3xz), \dfrac{d}{dy} (x^2 + 2y^2 + 3xz), \dfrac{d}{dz} (x^2 + 2y^2 + 3xz)\]

\[= (2x + 3z, 4y, 3x)\]

After that, **finding** $\nabla f(x_0)$:

\[\nabla f(1, 2, \dfrac{1}{3}) = (2 + 3 \dfrac{1}{3}, 4(2), 3)\]

\[\nabla f(x_0) = (3, 8, 3)\]

Here, plugging the **expressions** in the **formula**:

\[0=(3, 8, 3). (x-1, y-2, z – \dfrac{1}{3})\]

\[0=(3(x-1)+ 8(y-2) + 3(z – \dfrac{1}{3}))\]

\[0=(3x -3 + 8y-16 +3z – 1)\]

\[3x + 8y + 3z=20\]

**Part b:**

\[f(x,y,z) = y^2 – x^2, x_0=(1, 2, 8)\]

\[f(x_0) = 2^2 – 1^2=3\]

\[k=3\]

**Calculating** $ \nabla f(x)$:

\[\nabla f(x)=(\dfrac{d}{dx}(y^2 – x^2) , \dfrac{d}{dy} (y^2 – x^2), \dfrac{d}{dz} (y^2 – x^2) \]

\[= (-2x, 2y, 0)\]

After that, **finding** $ \nabla f(x_0)$:

\[\nabla f(1, 2, 8) = (-2 , 2(2), 0)\]

\[\nabla f(x_0) = (-2, 4, 0)\]

Again, plugging the **expressions** in the **formula**:

\[0 = (-2, 4, 0). (x-1, y-2, z – 8) = -2(x-1)+ 4(y-2) + 0(z – 8)\]

\[0 = (-2x +2 + 4y-8)\]

\[2y-x = 3\]

## Numerical Answer

**Part a:** $3x + 8y + 3z = 20$ is the **plane** **tangent** to the **surface** $x^2 + 2y^2 +3xz =1$ at the **point** $(1,2,\dfrac{1}{3})$.

**Part b:** $2y-x = 3$ is the **plane** **tangent** to the **surface** $y^2 -x^2 = 3$ at the **point** $(1,2,8)$.

## Example

Find the **plane** **tangent** to the given surface at the indicated **point**. $xyz = 1$**, **at the point $(1,1,1)$.

\[f(x,y,z) = (xyz), x_0 = (1, 1, 1)\]

\[f(x_0) = k = 1\]

Now **calculating** $ \nabla f(x)$:

\[\nabla f(x) = (\dfrac{d}{dx}(xyz) , \dfrac{d}{dy} (xyz), \dfrac{d}{dz} (xyz)\]

\[= (yz, xz, xy)\]

After that, **finding** $ \nabla f(x_0)$:

\[\nabla f(1, 1, 1) = (1 , 1, 1)\]

\[\nabla f(x_0) = (1, 1, 1)\]

Here, plugging the **expressions** in the **formula**:

\[0 = (1, 1, 1). (x-1, y-1, z – 1) = 1(x-1)+ 1(y-1) + 1(z – 1)\]

\[x+y+z=3\