Section 14.6: Directional Derivatives & Gradient Vector

Recall that if \(z = f(x, y)\), then the partial derivatives \(f_x\) and \(f_y\) are defined as

\(f_x(x_0, y_0) = \lim_{h\to0} \frac{f(x_0 + h, y_0) - f(x_0, y_0)}{h}\)

\(f_y(x_0, y_0) = \lim_{h\to0} \frac{f(x_0, y_0 + h) - f(x_0, y_0)}{h}\)

and represent the rates of change of \(z\) in the x- and y-directions, that is, in the directions of the unit vectors i and j.

Suppose that we now wish to find the rate of change of \(z\) at \((x_0, y_0)\) in the direction of an arbitrary unit vector \(\mathbf{u} = \langle a, b \rangle\). To do this we consider the surface S with the equation \(z = f(x, y)\) (the graph of f) and we let \(z_0 = f(x_0, y_0)\).

Then the point \(P(x_0, y_0, z_0)\) lies on S. The vertical plane that passes through P in the direction of u intersects S in a curve C. The slope of the tangent line T to C at the point P is the rate of change of z in the direction of u.

If \(Q(x, y, z)\) is another point on C and P’, Q’ are the projections of P, Q onto the xy-plane, then the vector \(\vec{P'Q'}\) is parallel to u and so

\(\vec{P'Q'} = h\mathbf{u} = \langle ha, hb \rangle\)

for some scalar h. Therefore \(x - x_0 = ha\), \(y - y_0 = hb\), so \(x = x_0 + ha\), \(y = y_0 + hb\), and

\(\frac{\Delta z}{h} = \frac{z - z_0}{h} = \frac{f(x_0 + ha, y_0 + hb) - f(x_0, y_0)}{h}\)

If we take the limit as \(h \to 0\), we obtain the rate of change of z (with respect to distance) in the direction of u, which is called the directional derivative of f in the direction of u.

Definition The directional derivative of \(f\) at \((x_0, y_0)\) in the direction of a unit vector \(\mathbf{u} = \langle a, b \rangle\) is

\(D_{\mathbf{u}}f(x_0, y_0) = \lim_{h\to0} \frac{f(x_0 + ha, y_0 + hb) - f(x_0, y_0)}{h}\)

if this limit exists.

Theorem If f is a differentiable function of x and y, then f has a directional derivative in the direction of any unit vector \(\mathbf{u} = \langle a, b \rangle\) and

\[ D_{\mathbf{u}}f(x, y) = f_x(x, y)a + f_y(x, y)b \]

PROOF If we define a function g of the single variable h by

\(g(h) = f(x_0 + ha, y_0 + hb)\)

then, by the definition of a derivative, we have

\(g'(0) = \lim_{h\to0} \frac{g(h) - g(0)}{h} = \lim_{h\to0} \frac{f(x_0 + ha, y_0 + hb) - f(x_0, y_0)}{h} = D_{\mathbf{u}}f(x_0, y_0)\)

On the other hand, we can write \(g(h) = f(x, y)\), where \(x = x_0 + ha\), \(y = y_0 + hb\), so the Chain Rule gives

\(g'(h) = \frac{\partial f}{\partial x}\frac{dx}{dh} + \frac{\partial f}{\partial y}\frac{dy}{dh} = f_x(x, y)a + f_y(x, y)b\)

If we now put \(h=0\), then \(x = x_0\), \(y = y_0\), and

\(g'(0) = f_x(x_0, y_0)a + f_y(x_0, y_0)b\)

Comparing the equations for \(g'(0)\), we see that

\(D_{\mathbf{u}}f(x_0, y_0) = f_x(x_0, y_0)a + f_y(x_0, y_0)b\)

If the unit vector u makes an angle \(\theta\) with the positive x-axis, then we can write \(\mathbf{u} = \langle \cos\theta, \sin\theta \rangle\) and the formula in Theorem 3 becomes

\(D_{\mathbf{u}}f(x, y) = f_x(x, y)\cos\theta + f_y(x, y)\sin\theta\)

EXAMPLE Find the directional derivative \(D_{\mathbf{u}}f(x, y)\) if

\(f(x, y) = x^3 - 3xy + 4y^2\)

and u is the unit vector given by angle \(\theta = \pi/6\). What is \(D_{\mathbf{u}}f(1, 2)\)?

Definition If f is a function of two variables x and y, then the gradient of f is the vector function \(\nabla f\) defined by

\[ \nabla f(x, y) = \langle f_x(x, y), f_y(x, y) \rangle = \frac{\partial f}{\partial x}\mathbf{i} + \frac{\partial f}{\partial y}\mathbf{j} \]

EXAMPLE If \(f(x, y) = \sin x + e^{xy}\), then

\(\nabla f(x, y) = \langle f_x, f_y \rangle = \langle \cos x + ye^{xy}, xe^{xy} \rangle\)

and

\(\nabla f(0, 1) = \langle 2, 0 \rangle\)

With this notation for the gradient vector, we can rewrite the equation for the directional derivative of a differentiable function as

\[ D_{\mathbf{u}}f(x, y) = \nabla f(x, y) \cdot \mathbf{u} \]

This expresses the directional derivative in the direction of a unit vector u as the scalar projection of the gradient vector onto u.

EXAMPLE Find the directional derivative of the function \(f(x, y) = x^2y^3 - 4y\) at the point \((2, -1)\) in the direction of the vector \(\mathbf{v} = 2\mathbf{i} + 5\mathbf{j}\).

For functions of three variables we can define directional derivatives in a similar manner. Again \(D_{\mathbf{u}}f(x, y, z)\) can be interpreted as the rate of change of the function in the direction of a unit vector u.

Definition The directional derivative of f at \((x_0, y_0, z_0)\) in the direction of a unit vector \(\mathbf{u} = \langle a, b, c \rangle\) is

\(D_{\mathbf{u}}f(x_0, y_0, z_0) = \lim_{h\to0} \frac{f(x_0 + ha, y_0 + hb, z_0 + hc) - f(x_0, y_0, z_0)}{h}\)

if this limit exists.

Gradients

For a function f of three variables, the gradient vector, denoted by \(\nabla f\) or grad f, is

\[ \nabla f = \langle f_x, f_y, f_z \rangle = \frac{\partial f}{\partial x}\mathbf{i} + \frac{\partial f}{\partial y}\mathbf{j} + \frac{\partial f}{\partial z}\mathbf{k} \]

Then, just as with functions of two variables, the formula for the directional derivative can be rewritten as

\[ D_{\mathbf{u}}f(x, y, z) = \nabla f(x, y, z) \cdot \mathbf{u} \]

EXAMPLE If \(f(x, y, z) = x \sin(yz)\), (a) find the gradient of f and (b) find the directional derivative of f at \((1, 3, 0)\) in the direction of \(\mathbf{v} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}\).

Suppose we have a function f of two or three variables and we consider all possible directional derivatives of f at a given point. These give the rates of change of f in all possible directions. We can then ask the questions: in which of these directions does f change fastest and what is the maximum rate of change? The answers are provided by the following theorem.

Theorem Suppose f is a differentiable function of two or three variables. The maximum value of the directional derivative \(D_{\mathbf{u}}f(\mathbf{x})\) is \(|\nabla f(\mathbf{x})|\) and it occurs when u has the same direction as the gradient vector \(\nabla f(\mathbf{x})\).

PROOF From the equations we have

\(D_{\mathbf{u}}f = \nabla f \cdot \mathbf{u} = |\nabla f||\mathbf{u}|\cos\theta = |\nabla f|\cos\theta\)

where \(\theta\) is the angle between \(\nabla f\) and u. The maximum value of \(\cos\theta\) is 1 and this occurs when \(\theta = 0\). Therefore the maximum value of \(D_{\mathbf{u}}f\) is \(|\nabla f|\) and it occurs when \(\theta = 0\), that is, when u has the same direction as \(\nabla f\).

EXAMPLE (a) If \(f(x, y) = xe^y\), find the rate of change of f at the point P(2, 0) in the direction from P to Q(1/2, 2). (b) In what direction does f have the maximum rate of change? What is this maximum rate of change?

EXAMPLE 7 Suppose that the temperature at a point (x, y, z) in space is given by \(T(x, y, z) = 80/(1 + x^2 + 2y^2 + 3z^2)\), where T is measured in degrees Celsius and x, y, z in meters. In which direction does the temperature increase fastest at the point (1, 1, -2)? What is the maximum rate of increase?

Suppose S is a surface with equation \(F(x, y, z) = k\), that is, it is a level surface of a function F of three variables, and let \(P(x_0, y_0, z_0)\) be a point on S. Let C be any curve that lies on the surface S and passes through the point P. The curve C is described by a continuous vector function \(\mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle\). Let \(t_0\) be the parameter value corresponding to P; that is, \(\mathbf{r}(t_0) = \langle x_0, y_0, z_0 \rangle\). Since C lies on S, any point \((x(t), y(t), z(t))\) must satisfy the equation of S, that is,

\(F(x(t), y(t), z(t)) = k\)

If x, y, and z are differentiable functions of t and F is also differentiable, then we can use the Chain Rule to differentiate both sides of the equation as follows:

\(\frac{\partial F}{\partial x}\frac{dx}{dt} + \frac{\partial F}{\partial y}\frac{dy}{dt} + \frac{\partial F}{\partial z}\frac{dz}{dt} = 0\)

But, since \(\nabla F = \langle F_x, F_y, F_z \rangle\) and \(\mathbf{r}'(t) = \langle x'(t), y'(t), z'(t) \rangle\), the equation can be written in terms of a dot product as

\(\nabla F \cdot \mathbf{r}'(t) = 0\)

In particular, when \(t = t_0\) we have \(\mathbf{r}(t_0) = \langle x_0, y_0, z_0 \rangle\), so

\(\nabla F(x_0, y_0, z_0) \cdot \mathbf{r}'(t_0) = 0\)

This equation says that the gradient vector at P, \(\nabla F(x_0, y_0, z_0)\), is perpendicular to the tangent vector \(\mathbf{r}'(t_0)\) to any curve C on S that passes through P. If \(\nabla F(x_0, y_0, z_0) \neq \mathbf{0}\), it is therefore natural to define the tangent plane to the level surface \(F(x, y, z) = k\) at \(P(x_0, y_0, z_0)\) as the plane that passes through P and has normal vector \(\nabla F(x_0, y_0, z_0)\). Using the standard equation of a plane, we can write the equation of this tangent plane as

\(F_x(x_0, y_0, z_0)(x - x_0) + F_y(x_0, y_0, z_0)(y - y_0) + F_z(x_0, y_0, z_0)(z - z_0) = 0\)

The normal line to S at P is the line passing through P and perpendicular to the tangent plane. The direction of the normal line is therefore given by the gradient vector \(\nabla F(x_0, y_0, z_0)\) and so, its symmetric equations are

\(\frac{x - x_0}{F_x(x_0, y_0, z_0)} = \frac{y - y_0}{F_y(x_0, y_0, z_0)} = \frac{z - z_0}{F_z(x_0, y_0, z_0)}\)

In the special case in which the equation of a surface S is of the form \(z = f(x, y)\), we can rewrite the equation as

\(F(x, y, z) = f(x, y) - z = 0\)

and regard S as a level surface (with \(k = 0\)) of F. Then

\(F_x(x_0, y_0, z_0) = f_x(x_0, y_0)\) \(F_y(x_0, y_0, z_0) = f_y(x_0, y_0)\) \(F_z(x_0, y_0, z_0) = -1\)

so the equation becomes

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

which is equivalent to our previous definition of a tangent plane. Thus our new, more general, definition of a tangent plane is consistent with the definition that was given for the special case.

EXAMPLE Find the equations of the tangent plane and normal line at the point \((-2, 1, -3)\) to the ellipsoid

\(\frac{x^2}{4} + y^2 + \frac{z^2}{9} = 3\)

A table of values for the wind-chill index \(W = f(T, v)\) is given in Exercise 14.3.3 on page 923. Use the table to estimate the value of \(D_u f(-20, 30)\), where \(u = (i + j)/\sqrt{2}\).

Find the directional derivative of \(f(x, y) = xy^3 - x^2\) at the point \((1, 2)\) in the direction indicated by the angle \(\theta = \pi/3\).

Find the directional derivative of \(f(x, y) = y \cos(xy)\) at the point \((0, 1)\) in the direction indicated by the angle \(\theta = \pi/4\).

Find the directional derivative of \(f(x, y) = \sqrt{2x + 3y}\) at the point \((3, 1)\) in the direction indicated by the angle \(\theta = -\pi/6\).

For \(f(x, y) = x/y\) at point \(P(2, 1)\) with vector \(u = \langle 5/13, 12/13 \rangle\): (a) Find the gradient of f. (b) Evaluate the gradient at the point P. (c) Find the rate of change of f at P in the direction of the vector u.

For \(f(x, y) = x^2 \ln y\) at point \(P(3, 1)\) with vector \(u = \langle 1/\sqrt{2}, -1/\sqrt{2} \rangle\): (a) Find the gradient of f. (b) Evaluate the gradient at the point P. (c) Find the rate of change of f at P in the direction of the vector u.

For \(f(x, y, z) = x^2yz - xyz^3\) at point \(P(2, -1, 1)\) with vector \(u = \langle 0, 4/5, -3/5 \rangle\): (a) Find the gradient of f. (b) Evaluate the gradient at the point P. (c) Find the rate of change of f at P in the direction of the vector u.

For \(f(x, y, z) = y^2e^{xyz}\) at point \(P(0, 1, -1)\) with vector \(u = \langle 3/13, 4/13, 12/13 \rangle\): (a) Find the gradient of f. (b) Evaluate the gradient at the point P. (c) Find the rate of change of f at P in the direction of the vector u.

Find the directional derivative of the function \(f(x, y) = e^x \sin y\) at the point \((0, \pi/3)\) in the direction of the vector \(v = \langle -6, 8 \rangle\).

Find the directional derivative of the function \(f(x, y) = \frac{x}{x^2 + y^2}\) at the point \((1, 2)\) in the direction of the vector \(v = \langle 3, 5 \rangle\).

Find the directional derivative of the function \(g(s, t) = s\sqrt{t}\) at the point \((2, 4)\) in the direction of the vector \(v = 2i - j\).

Find the directional derivative of the function \(g(u, v) = u^2e^{-v}\) at the point \((3, 0)\) in the direction of the vector \(v = 3i + 4j\).

Find the directional derivative of the function \(f(x, y, z) = \sqrt{x^2y + y^2z}\) at the point \((1, 2, 3)\) in the direction of the vector \(v = \langle 2, -1, 2 \rangle\).

Find the directional derivative of the function \(f(x, y, z) = xy^2 \tan^{-1}z\) at the point \((2, 1, 1)\) in the direction of the vector \(v = \langle 1, 1, 1 \rangle\).

Find the directional derivative of the function \(h(r, s, t) = \ln(3r + 6s + 9t)\) at the point \((1, 1, 1)\) in the direction of the vector \(v = 4i + 12j + 6k\).

Use the figure to estimate \(D_u f(2, 2)\).

Find the directional derivative of \(f(x, y) = \sqrt{xy}\) at \(P(2, 8)\) in the direction of \(Q(5, 4)\).

Find the directional derivative of \(f(x, y, z) = xy^2z^3\) at \(P(2, 1, 1)\) in the direction of \(Q(0, -3, 5)\).

Find the maximum rate of change of \(f(x, y) = 4y\sqrt{x}\) at the point \((4, 1)\) and the direction in which it occurs.

Find the maximum rate of change of \(f(s, t) = te^s\) at the point \((0, 2)\) and the direction in which it occurs.

Find the maximum rate of change of \(f(x, y) = \sin(xy)\) at the point \((1, 0)\) and the direction in which it occurs.

Find the maximum rate of change of \(f(x, y, z) = x \ln(yz)\) at the point \((1, 2, 1/2)\) and the direction in which it occurs.

Find the maximum rate of change of \(f(x, y, z) = x/(y + z)\) at the point \((8, 1, 3)\) and the direction in which it occurs.

Find the maximum rate of change of \(f(p, q, r) = \arctan(pqr)\) at the point \((1, 2, 1)\) and the direction in which it occurs.

1. Show that a differentiable function \(f\) decreases most rapidly at x in the direction opposite to the gradient vector, that is, in the direction of \(-\nabla f(\mathbf{x})\).
2. Use the result of part (a) to find the direction in which the function \(f(x, y) = x^4y - x^2y^3\) decreases fastest at the point \((2, -3)\).

Find the directions in which the directional derivative of \(f(x, y) = x^2 + \sin(xy)\) at the point \((1, 0)\) has the value 1.

Find all points at which the direction of fastest change of the function \(f(x, y) = x^2 + y^2 - 2x - 4y\) is \(\mathbf{i} + \mathbf{j}\).

Near a buoy, the depth of a lake at the point with coordinates \((x, y)\) is \(z = 200 + 0.02x^2 - 0.001y^3\), where x, y, and z are measured in meters. A fisherman in a small boat starts at the point \((80, 60)\) and moves toward the buoy, which is located at \((0, 0)\). Is the water under the boat getting deeper or shallower when he departs? Explain.

The temperature T in a metal ball is inversely proportional to the distance from the center of the ball, which we take to be the origin. The temperature at the point \((1, 2, 2)\) is 120°. (a) Find the rate of change of T at \((1, 2, 2)\) in the direction toward the point \((2, 1, 3)\). (b) Show that at any point in the ball the direction of greatest increase in temperature is given by a vector that points toward the origin.

The temperature at a point \((x, y, z)\) is given by \(T(x, y, z) = 200e^{-x^2 - 3y^2 - 9z^2}\) where T is measured in °C and x, y, z in meters. (a) Find the rate of change of temperature at the point \(P(2, -1, 2)\) in the direction toward the point \((3, -3, 3)\). (b) In which direction does the temperature increase fastest at P? (c) Find the maximum rate of increase at P.

Suppose that over a certain region of space the electrical potential V is given by \(V(x, y, z) = 5x^2 - 3xy + xyz\). (a) Find the rate of change of the potential at \(P(3, 4, 5)\) in the direction of the vector \(\mathbf{v} = \mathbf{i} + \mathbf{j} - \mathbf{k}\). (b) In which direction does V change most rapidly at P? (c) What is the maximum rate of change at P?

Suppose you are climbing a hill whose shape is given by the equation \(z = 1000 - 0.005x^2 - 0.01y^2\), where x, y, and z are measured in meters, and you are standing at a point with coordinates \((60, 40, 966)\). The positive x-axis points east and the positive y-axis points north. (a) If you walk due south, will you start to ascend or descend? At what rate? (b) If you walk northwest, will you start to ascend or descend? At what rate? (c) In which direction is the slope largest? What is the rate of ascent in that direction? At what angle above the horizontal does the path in that direction begin?

Let f be a function of two variables that has continuous partial derivatives and consider the points A(1, 3), B(3, 3), C(1, 7), and D(6, 15). The directional derivative of f at A in the direction of the vector \(\vec{AB}\) is 3 and the directional derivative at A in the direction of \(\vec{AC}\) is 26. Find the directional derivative of f at A in the direction of the vector \(\vec{AD}\).

Show that the operation of taking the gradient of a function has the given property. Assume that u and v are differentiable functions of x and y and that a, b are constants. (a) \(\nabla(au + bv) = a\nabla u + b\nabla v\) (b) \(\nabla(uv) = u\nabla v + v\nabla u\) (c) \(\nabla(\frac{u}{v}) = \frac{v\nabla u - u\nabla v}{v^2}\) (d) \(\nabla u^n = nu^{n-1}\nabla u\)

Sketch the gradient vector \(\nabla f(4, 6)\) for the function f whose level curves are shown. Explain how you chose the direction and length of this vector.

The second directional derivative of \(f(x, y)\) is \(D^2_{\mathbf{u}}f(x, y) = D_{\mathbf{u}}[D_{\mathbf{u}}f(x, y)]\) If \(f(x, y) = x^3 + 5x^2y + y^3\) and \(\mathbf{u} = \langle \frac{3}{5}, \frac{4}{5} \rangle\), calculate \(D^2_{\mathbf{u}}f(2, 1)\).

1. If \(\mathbf{u} = \langle a, b \rangle\) is a unit vector and f has continuous second partial derivatives, show that \(D^2_{\mathbf{u}}f = f_{xx}a^2 + 2f_{xy}ab + f_{yy}b^2\)
2. Find the second directional derivative of \(f(x, y) = xe^{2y}\) in the direction of \(\mathbf{v} = \langle 4, 6 \rangle\).

Find equations of (a) the tangent plane and (b) the normal line to the given surface \(2(x - 2)^2 + (y - 1)^2 + (z - 3)^2 = 10\) at the specified point \((3, 3, 5)\).

Find equations of (a) the tangent plane and (b) the normal line to the given surface \(x = y^2 + z^2 - 2\) at the specified point \((3, 1, -1)\).

Find equations of (a) the tangent plane and (b) the normal line to the given surface \(xy^2z^3 = 8\) at the specified point \((2, 2, 1)\).

Find equations of (a) the tangent plane and (b) the normal line to the given surface \(xy + yz + zx = 5\) at the specified point \((1, 2, 1)\).

Find equations of (a) the tangent plane and (b) the normal line to the given surface \(x + y + z = e^{xyz}\) at the specified point \((0, 0, 1)\).

Find equations of (a) the tangent plane and (b) the normal line to the given surface \(x^4 + y^4 + z^4 = 3x^2y^2z^2\) at the specified point \((1, 1, 1)\).

Use a computer to graph the surface, the tangent plane, and the normal line on the same screen for \(xy + yz + zx = 3\) at \((1, 1, 1)\). Choose the domain carefully so that you avoid extraneous vertical planes. Choose the viewpoint so that you get a good view of all three objects.

Use a computer to graph the surface, the tangent plane, and the normal line on the same screen for \(xyz = 6\) at \((1, 2, 3)\). Choose the domain carefully so that you avoid extraneous vertical planes. Choose the viewpoint so that you get a good view of all three objects.

If \(f(x, y) = xy\), find the gradient vector \(\nabla f(3, 2)\) and use it to find the tangent line to the level curve \(f(x, y) = 6\) at the point \((3, 2)\). Sketch the level curve, the tangent line, and the gradient vector.

If \(g(x, y) = x^2 + y^2 - 4x\), find the gradient vector \(\nabla g(1, 2)\) and use it to find the tangent line to the level curve \(g(x, y) = 1\) at the point \((1, 2)\). Sketch the level curve, the tangent line, and the gradient vector.

Show that the equation of the tangent plane to the ellipsoid \(x^2/a^2 + y^2/b^2 + z^2/c^2 = 1\) at the point \((x_0, y_0, z_0)\) can be written as \(\frac{xx_0}{a^2} + \frac{yy_0}{b^2} + \frac{zz_0}{c^2} = 1\)

Find the equation of the tangent plane to the hyperboloid \(x^2/a^2 + y^2/b^2 - z^2/c^2 = 1\) at \((x_0, y_0, z_0)\) and express it in a form similar to the one in Exercise 51.

Show that the equation of the tangent plane to the elliptic paraboloid \(z/c = x^2/a^2 + y^2/b^2\) at the point \((x_0, y_0, z_0)\) can be written as \(\frac{2xx_0}{a^2} + \frac{2yy_0}{b^2} = \frac{z + z_0}{c}\)

At what point on the ellipsoid \(x^2 + y^2 + 2z^2 = 1\) is the tangent plane parallel to the plane \(x + 2y + z = 1\)?

Are there any points on the hyperboloid \(x^2 - y^2 - z^2 = 1\) where the tangent plane is parallel to the plane \(z = x + y\)?

Show that the ellipsoid \(3x^2 + 2y^2 + z^2 = 9\) and the sphere \(x^2 + y^2 + z^2 - 8x - 6y - 8z + 24 = 0\) are tangent to each other at the point \((1, 1, 2)\). (This means that they have a common tangent plane at the point.)

Show that every plane that is tangent to the cone \(x^2 + y^2 = z^2\) passes through the origin.

Show that every normal line to the sphere \(x^2 + y^2 + z^2 = r^2\) passes through the center of the sphere.

Where does the normal line to the paraboloid \(z = x^2 + y^2\) at the point \((1, 1, 2)\) intersect the paraboloid a second time?

At what points does the normal line through the point \((1, 2, 1)\) on the ellipsoid \(4x^2 + y^2 + 4z^2 = 12\) intersect the sphere \(x^2 + y^2 + z^2 = 102\)?

Show that the sum of the x-, y-, and z-intercepts of any tangent plane to the surface \(\sqrt{x} + \sqrt{y} + \sqrt{z} = \sqrt{c}\) is a constant.

Show that the pyramids cut off from the first octant by any tangent planes to the surface \(xyz = 1\) at points in the first octant must all have the same volume.

Find parametric equations for the tangent line to the curve of intersection of the paraboloid \(z = x^2 + y^2\) and the ellipsoid \(4x^2 + y^2 + z^2 = 9\) at the point \((-1, 1, 2)\).

1. The plane \(y + z = 3\) intersects the cylinder \(x^2 + y^2 = 5\) in an ellipse. Find parametric equations for the tangent line to this ellipse at the point \((1, 2, 1)\).
2. Graph the cylinder, the plane, and the tangent line on the same screen.

Where does the helix \(\mathbf{r}(t) = \langle \cos(\pi t), \sin(\pi t), t \rangle\) intersect the paraboloid \(z = x^2 + y^2\)? What is the angle of intersection between the helix and the paraboloid? (This is the angle between the tangent vector to the curve and the tangent plane to the paraboloid.)

The helix \(\mathbf{r}(t) = \langle \cos(\pi t/2), \sin(\pi t/2), t \rangle\) intersects the sphere \(x^2 + y^2 + z^2 = 2\) in two points. Find the angle of intersection at each point.

1. Two surfaces are called orthogonal at a point of intersection if their normal lines are perpendicular at that point. Show that surfaces with equations \(F(x, y, z) = 0\) and \(G(x, y, z) = 0\) are orthogonal at a point P where \(\nabla F \neq \mathbf{0}\) and \(\nabla G \neq \mathbf{0}\) if and only if \(F_x G_x + F_y G_y + F_z G_z = 0\) at P

2. Use part (a) to show that the surfaces \(z^2 = x^2 + y^2\) and \(x^2 + y^2 + z^2 = r^2\) are orthogonal at every point of intersection. Can you see why this is true without using calculus?

1. Show that the function \(f(x, y) = \sqrt[3]{xy}\) is continuous and the partial derivatives \(f_x\) and \(f_y\) exist at the origin but the directional derivatives in all other directions do not exist.
2. Graph f near the origin and comment on how the graph confirms part (a).

Suppose that the directional derivatives of \(f(x, y)\) are known at a given point in two nonparallel directions given by unit vectors u and v. Is it possible to find \(\nabla f\) at this point? If so, how would you do it?

Show that if \(z = f(x, y)\) is differentiable at \(\mathbf{x}_0 = (x_0, y_0)\), then \(\lim_{\mathbf{x} \to \mathbf{x}_0} \frac{f(\mathbf{x}) - f(\mathbf{x}_0) - \nabla f(\mathbf{x}_0) \cdot (\mathbf{x} - \mathbf{x}_0)}{|\mathbf{x} - \mathbf{x}_0|} = 0\)

[Hint: Use Definition Directly]