# 14. Directional Derivatives and Gradients

## e. Tangent and Normal, Lines and Planes to Level Sets

Since the gradient vector is perpendicular to each level set of a function,
it is useful in finding

1. the tangent line to a level curve in \(\mathbb R^2\)
(this page)

2. the tangent plane to a level surface in \(\mathbb R^3\)
(the
next page)

3. the normal line to a level curve in \(\mathbb R^2\)
(the
subsequent page)

4. the normal line to a level surface in \(\mathbb R^3\)
(the
subsequent page)

## 1. Tangent Lines to Level Curves in \(\mathbb R^2\)

A curve in the plane is frequently defined as the graph of an equation. In that case, the curve can be regarded as the level set of a function \(F(x,y)\). To determine the tangent line to the curve at a point \(P=(a,b)\), we can find the normal vector at \(P\), which in turn, is the gradient of \(F\) at \(P\).

Find the line tangent to the curve \(x^3y^2=4\) at the point \((x,y)=(1,2)\).

Recall from the chapter on Lines and Planes that the normal equation for a line in \(\mathbb R^2\) is given by \[ \vec n\cdot X=\vec n\cdot P \] where \(P\) is a particular point on the line, \(\vec n\) is a normal vector to the line and \(X=(x,y)\) is a general point on the line. In the case of a tangent line to a level curve of a function \(F\), \(P\) is the point of tangency and \(\vec n\) is the normal to the level curve, i.e. the gradient of \(F\) at \(P\).

In this example, the function is \(F(x,y)=x^3y^2\), its gradient is \[ \vec\nabla F=\langle 3x^2y^2,2x^3y\rangle \] and the normal at \(P=(1,2)\) is \[ \vec n=\left.\vec\nabla F\right|_{(1,2)} =\langle 12,4\rangle \] So the normal equation of the line is \[ \vec n\cdot X =\vec n\cdot P \qquad \text{or} \qquad \langle 12,4\rangle\cdot(x,y) =\langle 12,4\rangle\cdot(1,2) \] or \[ 12x+4y=20 \qquad \text{or} \qquad 3x+y=5. \]

Find the line tangent to the ellipse \(\dfrac{x^2}{9}+\dfrac{y^2}{4}=2\) at the point \((x,y)=(3,2)\).

\(2x+3y=12\)

The gradient of the function \(f=\dfrac{x^2}{9}+\dfrac{y^2}{4}\) is \[ \vec\nabla f=\left\langle \dfrac{2x}{9},\dfrac{y}{2}\right\rangle \] Its value at \(P=(3,2)\) is the normal: \[ \vec n=\left.\vec\nabla f\right|_P =\left\langle \dfrac{2}{3},1\right\rangle \] So the tangent line \(\vec n\cdot X=\vec n\cdot P\) is \[ \dfrac{2}{3}x+y=\dfrac{2}{3}3+2=4 \] or equivalently, \[ 2x+3y=12 \]

It is important to distinguish between three (or four) methods of finding the tangent line to a curve.

- If the curve is given as the graph of the a function, \(y=f(x)\), then the tangent line at \(x=a\) is: \[ y=f_{\tan}(x)=f(a)+f'(a)(x-a) \]
- If the curve is given as a parametric curve, \(\vec r(t)=(x(t),y(t))\), then the tangent vector at \(t=a\) is \(\vec v(a)=(x'(a),y'(a))\) and the tangent line at \(\vec r(a)=(x(a),y(a))\) is: \[ \vec r_{\tan}(t)=\vec r(a)+t\,\vec v(a) \]
- If the curve is given as the graph of an equation \(F(x,y)=C\), (equivalently, the level curve of the function \(F(x,y)\) with value \(C\)), then the normal vector is \(\vec n=\left.\vec\nabla F\right|_{(a,b)}\) and the tangent line at \(P=(a,b)\) is: \[ \vec n\cdot X=\vec n\cdot P \]
- If the curve is given as the graph of an equation \(F(x,y)=C\), then an alternate method to find the tangent line at \((a,b)\) is to apply method (1) to the implicitly defined function \(y=f(x)\): Since the point is \((a,b)\), the function value is \(f(a)=b\). Using implicit differentiation the slope is \(f'(a)=\left.\dfrac{dy}{dx}\right|_{(a,b)}\). Then the tangent line at \((a,b)\) is: \[ y=f_{\tan}(x)=f(a)+f'(a)(x-a) \] This implicit differentiation method is the least useful of the four. In general, if the curve is a level set \(F(x,y)=C\), then it is much easier to compute the gradient, \(\vec\nabla F\), than to use implicit differentiation to compute the slope, \(\dfrac{dy}{dx}\).

Find the tangent line to the graph of the function \(y=f(x)=x^2\) at \((2,4)\).

\(y=4x-4\)

The derivative is \(f'(x)=2x\). At \(x=2\), the slope is \(m=4\). So the tangent line is \[ y=f_{\tan}(x)=f(2)+f'(2)(x-2)=4+4(x-2)=4x-4 \]

Find the tangent line to the parametric curve \(\vec r(t)=(t,t^2)\) at \((2,4)\).

\(x=2+t \qquad y=4+4t\)

The tangent vector at a general point is \(\vec v=\langle 1,2t\rangle\). The tangent vector at \(t=2\) is \(\vec v(2)=\langle 1,4\rangle\). So the tangent line is \[ X=(x,y)=\vec r_{\tan}(t)=\vec r(2)+t\,\vec v(2) =\langle 2,4\rangle+t\,\langle 1,4\rangle \] or \[ x=2+t \qquad y=4+4t \]

Find the tangent line to the level set of the function \(F(x,y)=y-x^2\) at \((2,4)\).

\(4x-y=4\)

The gradient is \(\vec\nabla F=\langle -2x,1\rangle\). The normal is \(\vec n=\left.\vec\nabla F\right|_{(2,4)}=\langle -4,1\rangle\). So the tangent line \(\vec n\cdot X=\vec n\cdot P\) is \[ -4x+y=-8+4=-4 \qquad \text{or} \qquad 4x-y=4 \]

Notice that these three exercises were all the same problem. Which method did you think was easiest? For more complex problems, different methods may have a distinct advantage, depending on the way the problem is stated.

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