Tangent Planes & Hyperplanes

🎯 Learning Objectives

• Find the equation of the tangent plane to $z = f(x,y)$ at a point
• Find the tangent hyperplane for $f: \mathbb{R}^n \to \mathbb{R}$
• Use the tangent plane for linear approximation
• Derive the tangent plane from the gradient

📋 Prerequisites

• Gradient (Week 9) — provides the normal vector
• Partial Derivatives (Week 9) — components of the gradient

1. Intuition: The Best Flat Approximation

2. Tangent Plane to $z = f(x,y)$

2.1 Formula

Definition (Tangent Plane). For $z = f(x,y)$, the tangent plane at $(a,b,f(a,b))$ is:

Or equivalently: $z = f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b)$. Example 1: Tangent plane

$f(x,y) = x^2 + y^2$ at $(1,2)$.

$f(1,2) = 5$, $f_x = 2x$, $f_x(1,2) = 2$, $f_y = 2y$, $f_y(1,2) = 4$.

Tangent plane: $z - 5 = 2(x-1) + 4(y-2)$, or $z = 2x + 4y - 5$.

2.2 Using the Gradient

3. Tangent Hyperplane

Example 2: Tangent hyperplane in ℝ³

$f(x,y,z) = x^2 + y^2 + z^2$ at $(1,1,1)$.

$\nabla f = (2x, 2y, 2z)$, $\nabla f(1,1,1) = (2,2,2)$. $f(1,1,1) = 3$.

Tangent hyperplane: $w - 3 = 2(x-1) + 2(y-1) + 2(z-1)$. Or $w = 2x + 2y + 2z - 3$.

4. Linear Approximation

Definition (Linearisation). The linearisation of $f$ at $\mathbf{a}$ is:

Example 3: Linear approximation

Approximate $\sqrt{(3.1)^2 + (4.2)^2}$ using $f(x,y) = \sqrt{x^2 + y^2}$ at $(3,4)$.

$f(3,4) = 5$. $f_x = \frac{x}{\sqrt{x^2+y^2}}$, $f_x(3,4) = \frac{3}{5}$. $f_y = \frac{y}{\sqrt{x^2+y^2}}$, $f_y(3,4) = \frac{4}{5}$.

$L(3.1, 4.2) = 5 + \frac{3}{5}(0.1) + \frac{4}{5}(0.2) = 5 + 0.06 + 0.16 = 5.22$.

Actual: $\sqrt{9.61 + 17.64} = \sqrt{27.25} \approx 5.2199$. Spot on!

5. Edge Cases & Gotchas

6. Common Pitfalls

❌ Pitfall 1: Forgetting the constant term

❌ Pitfall 2: Using unnormalised gradient for the plane equation

7. Formula Summary Table

8. 📝 Practice Questions

Q1: Tangent plane

Find tangent plane to $z = \sin(xy)$ at $(\pi/2, 1, 1)$.

Solution: $f_x = y\cos(xy)$, $f_x(\pi/2,1) = 1\cdot\cos(\pi/2) = 0$. $f_y = x\cos(xy)$, $f_y(\pi/2,1) = (\pi/2)\cos(\pi/2) = 0$. Tangent plane: $z - 1 = 0(x-\pi/2) + 0(y-1)$ ⇒ $z = 1$ (horizontal plane). Q2: Linear approximation

Approximate $e^{0.1}\cos(0.2)$ using $f(x,y) = e^x\cos y$ at $(0,0)$.

Solution: $f(0,0)=1$, $f_x = e^x\cos y$, $f_x(0,0)=1$, $f_y = -e^x\sin y$, $f_y(0,0)=0$. $L(0.1,0.2) = 1 + 1(0.1) + 0(0.2) = 1.1$. Actual: $e^{0.1}\cos(0.2) \approx 1.10517 \times 0.98007 \approx 1.083$. Not great — error due to second-order terms. Q3: Tangent plane from gradient

Find tangent plane to $z = x^2 - y^2$ at $(2,1,3)$ using gradient method.

Solution: $F(x,y,z) = z - x^2 + y^2 = 0$. $\nabla F = (-2x, 2y, 1)$, $\nabla F(2,1,3) = (-4, 2, 1)$. Plane: $(-4,2,1)\cdot(x-2, y-1, z-3) = 0$ ⇒ $-4(x-2) + 2(y-1) + (z-3) = 0$ ⇒ $-4x+8+2y-2+z-3=0$ ⇒ $z = 4x - 2y - 3$. Check: $f(2,1) = 3 = 4(2) - 2(1) - 3 = 8-2-3 = 3$ ✓. Q4: Multivariable linearisation

Linearise $f(x,y,z) = \sqrt{x^2 + y^2 + z^2}$ at $(3,4,0)$.

Solution: $f(3,4,0) = 5$. $\nabla f = \left(\frac{x}{\sqrt{x^2+y^2+z^2}}, \frac{y}{\sqrt{\dots}}, \frac{z}{\sqrt{\dots}}\right)$. $\nabla f(3,4,0) = \left(\frac{3}{5}, \frac{4}{5}, 0\right)$. $L(x,y,z) = 5 + \frac{3}{5}(x-3) + \frac{4}{5}(y-4) + 0(z-0) = \frac{3}{5}x + \frac{4}{5}y$. Q5: Error estimation

The radius and height of a cylinder are measured with errors. Use linearisation: $V = \pi r^2 h$. If $r=5$ (error 0.1), $h=10$ (error 0.1), estimate max error in $V$.

Solution: $V(5,10) = 250\pi$. $V_r = 2\pi rh$, $V_r(5,10) = 100\pi$. $V_h = \pi r^2$, $V_h(5,10) = 25\pi$. $\Delta V \approx |V_r|\Delta r + |V_h|\Delta h = 100\pi(0.1) + 25\pi(0.1) = 12.5\pi \approx 39.27$. Max error: about $12.5\pi$. Q6: Tangent plane to implicit surface

Find tangent plane to $x^2 + y^2 + z^2 = 14$ at $(1,2,3)$.

Solution: $F(x,y,z) = x^2 + y^2 + z^2 - 14$, $\nabla F = (2x, 2y, 2z)$, $\nabla F(1,2,3) = (2,4,6)$. Tangent plane: $2(x-1) + 4(y-2) + 6(z-3) = 0$ ⇒ $x + 2y + 3z = 14$. Q7: Differentiability check

Show $f(x,y) = x^2 + y^2$ is differentiable at $(1,2)$ by finding its linearisation and verifying the error goes to 0.

Solution: $L(x,y) = 5 + 2(x-1) + 4(y-2) = 2x + 4y - 5$. Error: $f(x,y) - L(x,y) = (x^2+y^2) - (2x+4y-5) = (x^2-2x+1) + (y^2-4y+4) = (x-1)^2 + (y-2)^2$. As $(x,y)\to(1,2)$, the error is $O(\|(x-1,y-2)\|^2)$, which goes to 0 faster than $\|(x-1,y-2)\|$. ✓ Q8: Tangent plane for $f(x,y)$ at critical point

$f(x,y) = x^2 + y^2$ at $(0,0,0)$. What's special about the tangent plane?

Solution: $\nabla f(0,0) = (0,0)$. Tangent plane: $z - 0 = 0(x-0) + 0(y-0)$ ⇒ $z = 0$. At a critical point ($\nabla f = 0$), the tangent plane is horizontal. Q9: Chain rule and tangent plane

If $z = f(x,y)$ and $x = g(t)$, $y = h(t)$, how does the tangent plane relate to $\frac{dz}{dt}$?

Solution: The tangent plane gives $\Delta z \approx f_x\Delta x + f_y\Delta y$. Dividing by $\Delta t$ and taking the limit gives $\frac{dz}{dt} = f_x\frac{dx}{dt} + f_y\frac{dy}{dt}$ — the chain rule. Q10: Total derivative

The total derivative $df$ is $df = f_x dx + f_y dy$. Show this corresponds to the linearisation.

Solution: The total derivative represents the linear part of the change in $f$: $\Delta f \approx f_x \Delta x + f_y \Delta y = \nabla f \cdot (\Delta x, \Delta y)$. This is exactly the linearisation $L(x,y) - f(a,b) = \nabla f(a,b) \cdot (x-a, y-b)$.

🔗 Cross-References

• Next topic: Critical Points
• Week 11 (Hessian): Second-order Taylor expansion refines the linear approximation
• BSMA1001 (Maths 1): Tangent line in single-variable calculus Join Discord Previous9.3 Limits & ContinuityNext10.2 Critical Points