Math 21a Tangent Lines and Planes Fall, What do we know about the gradient f? Tangent Lines to Curves in the Plane.

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1 Math 21a Tangent Lines and Planes Fall, 2016 What do we know about the gradient f? Tangent Lines to Curves in the Plane.

2 1. For each of the following curves, find the tangent line to the curve at the point (1, 1): (a) The level curve f(x, y) = 2 for f(x, y) = x 2 + y 2 (b) The level curve y 2x 2 log x = 1 (c) The level curve f(x, y) = e for f(x, y) = x 2 e y (d) The graph y = x 2 (e) The level curve f(x, y) = 2 for f(x, y) = xy 2 3x 3 y 5 (f) The parametric curve r(t) = t 2 + 1, e t

3 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) x 2 + y 2 + z 2 = 3 at (1, 1, 1) (b) xyz x 2 + yz = 11 at (1, 2, 3) (c) x 3 2y sin z = 8 at (2, 1, 0) (d) z = x 2 + 3xy + y at (2, 1, 2)

4 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) = x 2 + y 2. (b) Find the tangent plane to the graph z = f(x, y) at the point (1, 1, 2), where f(x, y) = x 2 + y 2. (c) Find the tangent plane to the parametric surface r(u, v) = 2 cos u, 2 sin u, v at the point (u, v) = ( π 4, 2).

5 Tangent Lines and Planes Answers and Solutions 1. (a) x + y = 2 i. As f = 2x, 2y, we have f(1, 1) = 2, 2. 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 y (x) = f x f y = 2x 2y = x y. So the slope at x = 1 is y (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) = 2 cos t, 2 sin t. Then r(π/4) = 1, 1. As r (t) = 1 2 sin t, 2 cos t, we get r (π/4) = 1, 1. So x, y = 1, 1 +t 1, 1 gives a parametrization of the tangent line. As x, y = 1, 1 +t 1, 1 = 1 t, 1+t, we have x 1 1 = t = y 1, namely, (x 1) = y 1 or x + y = 2. 1 (b) Write y = 2x 2 + log x 1. As y = 4x + 1 x, y (1) = 5. So y = 5x 4 gives the tangent line. (c) As f = 2xe y, x 2 e y, we have f(1, 1) = 2e, e. 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. (d) y = 2x 1 (e) As f = y 2 9x 2 y 5, 2xy 15x 3 y 4, we have f(1, 1) = 8, 13. 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. (f) Note that r(0) = 1, 1. A direction vector of the tangent line is given by the velocity vector r (0). As r (t) = 2t, e t, so r (0) = 0, 1. Hence x, y = 1, 1 + t 0, e gives a parametrization of the tangent line. This is the same as the line defined by x 1 = (a) As F = 2x, 2y, 2z, we have F (1, 1, 1) = 2, 2, 2. 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 x + y + z = 3 gives the tangent plane. (b) As F = yz 2x, xz + z, xy + y, we have F (1, 2, 3) = 4, 6, 4. 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 = 3x 2, 2 sin z, 2y cos z, we have F (2, 1, 0) = 12, 0, 2. 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) = x 2 +3xy+ y 3 +1 z (the right hand side minus the left hand side). Then the surface is defined by F (x, y, z) = 0. As F = 2x+3y, 3x+3y 2, 1, we have F (2, 1, 2) = 1, 9, 1. 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.

6 ii. The equation is of graph type: if you put f(x, y) = x 2 + 3xy + y 3 + 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) + f x (2, 1) ( x 2 ) + f y (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 = Parts (a) and (b) are intended to clarify our method. (a) As F = 2x, 2y, 0, we have F (1, 1, 2) = 2, 2, 0. 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) = x 2 + y 2 z. Then the surface is defined by F (x, y, z) = 0 and our method works. As F = 2x, 2y, 1, we have F (1, 1, 2) = 2, 2, 1. 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) = x 2 + y 2 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) = 1, 1, 2. (As r(u, v) is a parametrization of the surface x 2 +y 2 = 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 r u (u, v) = 2 sin u, 2 cos u, 0, we have r u (π/2, 2) = 1, 1, 0. This is a direction vector of the tangent line to the grid curve r(u, 2) at u = π/2. As r v (u, v) = 0, 0, 1, we have r v (π/2, 2) = 0, 0, 1. This is a direction vector of the tangent line to the grid curve r(π/4, v) at v = 2. So x, y, z = 1, 1, 2 + u 1, 1, 0 + v 0, 0, 1 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 a, b, c is a normal vector to the plane. We know such a vector, namely, 1, 1, 0 0, 0, 1! As 1, 1, 0 0, 0, 1 = 1, 1, 0, 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)).

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