Tangent Lines and Planes Fall, 2015

Math 21a Tangent Lines and Planes Fall, 2015

The goal for today’s class is to use the gradient of functions to ﬁnd tangent lines to curves in the xy-plane or tangent planes to surfaces in the xyz-space. Tangent Lines to Curves in the Plane.

1. For each of the following curves, ﬁnd the tangent line to the curve at the point (1, 1): (a) The graph y = x2 (b) The level curve y − 2x2 − log x = −1

(e) The level curve f(x, y) = e (f) The level curve f(x, y) = −2 for f(x, y) = x2ey for f(x, y) = xy2 − 3x3y5 Tangent Planes to Surfaces in the Space.

2. For each of the following equations, ﬁnd the tangent plane to the surface deﬁned by the equation at the given point. (a) x2 + y2 + z2 = 3 at (1, 1, 1) (b) xyz − x2 + yz = 11 at (1, 2, 3)

(c) x3 − 2y sin z = 8 at (2, 1, 0) (d) z = x2 + 3xy + y3 + 1 at (2, −1, −2) 3. (a) Find the tangent plane to the level surface F (x, y, z) = 2 at the point (1, 1, 2), where F (x, y, z) = x2 + y2.

(b) Find the tangent plane to the graph z = f(x, y) at the point (1, 1, 2), where f(x, y) = x2 + y2.

√ √ (c) Find the tangent plane to the parametric surface ~r(u, v) = h 2 cos u, 2 sin u, vi at the π point (u, v) = ( 4 , 2). More on Gradients.

Gradient vectors are orthogonal to level curves and level surfaces.

4. (a) Find the normal line to the level curve f(x, y) = −2 for f(x, y) = xy2 − 3x3y5 passing through the point (1, 1).

(b) Find the normal line to the surface deﬁned by xyz − x2 + yz = 11 passing through the point (1, 2, 3). Find the distance between the point (0, 0, 0) and the tangent plane to the surface at (1, 2, 3). Tangent Lines and Planes – Answers and Solutions

Tangent Lines to Curves in the Plane. We know how to ﬁnd tangent lines to a graph y = g(x) or a parametric curve ~r(t). So let’s discuss the way to ﬁnd tangent lines to a level curve f(x, y) = c.

Let’s take f(x, y) = x2 + y2 and c = 2, and ﬁnd the tangent line at the point (1, 1). The tangent line is the line which best approximates the curve at (1, 1). So let’s ﬁrst approximate f(x, y); this is the linear approximation! The linear approximation L(x, y) of f(x, y) at (1, 1) is L(x, y) = f(1, 1) + fx(1, 1) x − 1 + fy(1, 1) y − 1 = 2 + 2(x − 1) + 2(y − 1).

As our curve is deﬁned by f(x, y) = 2, the tangent line at (1, 1), which best approximates the curve at (1, 1), is given by L(x, y) = 2, i.e. 2 + 2(x − 1) + 2(y − 1) = 2 or 2x + 2y = 2 or x + y + 1.

There is a bit easier way. We know the tangent line should be written in the form ax + by = d, and we know a candidate for ha, bi: this is the gradient vector ∇f = hfx, fyi! As ∇f = h2x, 2yi, we have ∇f(1, 1) = h2, 2i. So our equation should be 2x + 2y = d. Plugging in (x, y) = (1, 1), we get d = 4. So an equation for the tangent line is 2x + 2y = 4 or x + y = 2.

Easy Method to ﬁnd tangent lines to level curves.

The tangent line to a level curve f(x, y) = c at the point (x0, y0) is written in the form

ax + by = d where ha, bi = ∇f(x0, y0).

We can then ﬁnd the constant d by plugging in (x, y) = (x0, y0).

We remark that this expression shows that ∇f(x0, y0) gives a normal vector to the level curve in the plane, i.e., ∇f is orthogonal to the level curve. Tangent Planes to Surfaces in the Space. First let’s consider tangent planes to a level surface F (x, y, z) = c. As in the tangent line case, tangent planes are given by L(x, y, z) = c where L(x, y, z) is the linearization of F (x, y, z) at the given point. From this we get the following:

Easy Method to ﬁnd tangent planes to level surfaces.

The tangent line to a level surface F (x, y, z) = c at the point (x0, y0, z0) is written in the form

ax + by + cz = d where ha, b, ci = ∇F (x0, y0, z0).

We can then ﬁnd the constant d by plugging in (x, y, z) = (x0, y0, z0).

We remark that this expression shows that ∇F (x0, y0, z0) gives a normal vector to the level surface in the space, i.e., ∇F is orthogonal to the level surface.

Next let’s discuss the tangent plane to a parametric surface ~r(u, v) at (u, v) = (u0, v0). In this case, we use the following observation: a plane is determined by a point on the plane and two vectors parallel to the plane. Let’s consider parametric curves ~r(u, v0) and ~r(u0, v)( u0 and v0 are constants). Then we can ﬁnd the tangent lines to the curves, and they will be on the tangent plane. So the directional vectors to these lines are vectors parallel to the tangent plane! We know what the directional vectors are. They are ~ru(u0, v0) and ~rv(u0, v0). So the tangent planes is parametrically given by hx, y, zi = ~r(u0, v0) + u~ru(u0, v0) + v~rv(u0, v0).

If we want to find a normal vector and write the tangent plane in the form ax + by + cz = d, then we can just compute ~ru(u0, v0) × ~rv(u0, v0). This will give ha, b, ci! See Solutions for Problem 3 (c) for detail. 1. (a) y = 2x − 1 2 0 1 0 (b) Write y = 2x + log x − 1. As y = 4x + x , y (1) = 5. So y = 5x − 4 gives the tangent line. (c) Note that ~r(0) = h1, 1i. A direction vector of the tangent line is given by the velocity vector ~r 0(0). As ~r 0(t) = h2t, eti, so ~r 0(0) = h0, 1i. Hence hx, yi = h1, 1i + th0, ei gives a parametrization of the tangent line. This is the same as the line defined by x − 1 = 0. (d) x + y = 2 i. As ∇f = h2x, 2yi, we have ∇f(1, 1) = h2, 2i. So the equation can be written in the form 2x + 2y = d. As (1, 1) is on the line, we get d = 4. So 2x + 2y = 4 or x + y = 2 gives the tangent line. ii. By implicit differentiation, we have f 2x x y0(x) = − x = − = − . fy 2y y So the slope at x = 1 is y0(1) = −1. From this we get y = −x + 2. This is the same as x + y = 2. √ √ iii. We parametrize√ the√ curve by ~r(t) = h 2 cos t, 2 sin ti. Then ~r(π/4) = h1, 1i. As ~r 0(t) = h1− 2 sin t, 2 cos ti, we get ~r 0(π/4) = h−1, 1i. So hx, yi = h1, 1i+th−1, 1i gives a parametrization of the tangent line. As hx, yi = h1, 1i+th−1, 1i = h1−t, 1+ti, we have x − 1 y − 1 = t = , namely, − (x − 1) = y − 1 or x + y = 2. −1 1 (e) As ∇f = h2xey, x2eyi, we have ∇f(1, 1) = h2e, ei. So the equation can be written in the form 2ex + ey = d. As (1, 1) is on the line, we get d = 3e. So 2ex + ey = 3e or 2x + y = 3 gives the tangent line. (f) As ∇f = hy2 − 9x2y5, 2xy − 15x3y4i, we have ∇f(1, 1) = h−8, −13i. So the equation can be written in the form −8x − 13y = d. As (1, 1) is on the line, we get d = −21. So −8x − 13y = −21 or 8x + 13y = 21 gives the tangent line. 2. (a) As ∇F = h2x, 2y, 2zi, we have ∇F (1, 1, 1) = h2, 2, 2i. So the equation can be written in the form 2x+2y +2z = d. As (1, 1, 1) is on the surface, we get d = 6. So 2x+2y +2z = 6 or z + y + z = 3 gives the tangent plane. (b) As ∇F = hyz − 2x, xz + z, xy + yi, we have ∇F (1, 2, 3) = h4, 6, 4i. So the equation can be written in the form 4x + 6y + 4z = d. As (1, 2, 3) is on the surface, we get d = 28. So 4x + 6y + 4z = 28 or 2x + 3y + 2z = 14 gives the tangent plane. (c) As ∇F = h3x2, −2 sin z, −2y cos zi, we have ∇F (2, 1, 0) = h12, 0, −2i. So the equation can be written in the form 12x − 2z = d. As (2, 1, 0) is on the surface, we get d = 24. So 12x − 2z = 24 or 6x − z = 12 gives the tangent plane. (d) i. First let’s rewrite the equation in the form F (x, y, z) = c. Put F (x, y, z) = x2 +3xy+ y3+1−z (the right hand side minus the left hand side). Then the surface is defined by F (x, y, z) = 0. As ∇F = h2x+3y, 3x+3y2, −1i, we have ∇F (2, −1, −2) = h1, 9, −1i. So the equation can be written in the form x + 9y − z = d. As (2, −1, −2) is on the surface, we get d = −5. So x + 9y − z = −5 gives the tangent plane. ii. The equation is of graph type: if you put f(x, y) = x2 + 3xy + y3 + 1, the equation is z = f(x, y). We want to approximate this equation by a linear equation. So we use the linear approximation L(x, y) of f(x, y) at (x, y) = (2, −1): L(x, y) = f(2, −1) + fx(2, −1) x − 2 + fy(2, −1) y − (−1) = −2 + 1(x − 2) + 9(y + 1) = x + 9y + 5.

Then the tangent plane to z = f(x, y) is given by the equation z = L(x, y), i.e., z = x + 9y + 5. This is the same as x + 9y − z = −5.

3. Parts (a) and (b) are intended to clarify our method.

(a) As ∇F = h2x, 2y, 0i, we have ∇F (1, 1, 2) = h2, 2, 0i. So the equation can be written in the form 2x + 2y = d. As (1, 1, 2) is on the surface, we get d = 4. So 2x + 2y = 4 or x + y = 2 gives the tangent plane. Note that F is a function of three variables x, y and z. So ∇F has three components, and it is a normal vector to the surface. (b) Put F (x, y, z) = x2 + y2 − z. Then the surface is defined by F (x, y, z) = 0 and our method works. As ∇F = h2x, 2y, −1i, we have ∇F (1, 1, 2) = h2, 2, −1i. So the equation can be written in the form 2x + 2y − z = d. As (1, 1, 2) is on the surface, we get d = 2. So 2x + 2y − z = 2 gives the tangent plane. A typical confusion arises when you just compute ∇f. This has only two components! We need three numbers a, b and c for ax + by + cz = d!Of course, one of the right ways is to consider F (x, y, z) = x2 + y2 − z.. Or as in Problem 2 (d), we can use lineariztion of f(x, y). You will find L(x, y) = 2 + 2(x − 1) + 2(y − 1) = 2x + 2y − 2. So z = L(x, y) = 2x + 2y − 2 is an equation for the tangent plane. (c) Note that ~r(π/4, 2) = h1, 1, 2i. (As ~r(u, v) is a parametrization of the surface x2 +y2 = 2, we are trying to find the same tangent plane as in (a).) First we consider grid curves ~r(u, 2) and ~r(π/4, v) to find√ two vectors√ (their direction vectors) parallel to the tangent plane. As ~ru(u, v) = h− 2 sin u, 2 cos u, 0i, we have ~ru(π/2, 2) = h−1, 1, 0i. This is a direction vector of the tangent line to the grid curve ~r(u, 2) at u = π/2. As ~rv(u, v) = h0, 0, 1i, we have ~rv(π/2, 2) = h0, 0, 1i. This is a direction vector of the tangent line to the grid curve ~r(π/4, v) at v = 2. So hx, y, zi = h1, 1, 2i + uh−1, 1, 0i + vh0, 0, 1i gives a parametrization of the tangent plane. We can also find an equation of the form ax + by + cz = d. For this let’s recall ha, b, ci is a normal vector to the plane. We know such a vector, namely, h−1, 1, 0i × h0, 0, 1i! As h−1, 1, 0i × h0, 0, 1i = h1, 1, 0i, the equation can be written in the form x + y = d. As (1, 1, 2) is on the tangent plane, we get d = 2. So x + y = 2 gives the tangent plane (and is the same as the answer for (a)).

4. (a) The gradient vector ∇f(1, 1) gives a direction vector of the normal line. As ∇f = hy2 − 9x2y5, 2xy − 15x3y4i, we get ∇f(1, 1) = h−8, −13i. So the tangent line is given parametrically by ~r(t) = h1, 1i + th−8, −13i. (b) The gradient vector ∇F (1, 2, 3) gives a direction vector of the normal line. As ∇F = hyz − 2x, xz + z, xy + yi, we get ∇F (1, 2, 3) = h4, 6, 4i. So the tangent line is given parametrically by ~r(t) = h1, 2, 3i + th4, 6, 4i. The distance is √ |h4, 6, 4i · h1, 2, 3i| 28 14 17 | proj h1, 2, 3i| = = √ = . h4,6,4i |h4, 6, 4i| 68 17