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 (c) The parametric curve (d) The level curve f(x; y) = 2 ~r(t) = ht2 + 1; eti for f(x; y) = x2 + y2 (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. p p (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 rf = hfx; fyi! As rf = h2x; 2yi, we have rf(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 = rf(x0; y0): We can then find the constant d by plugging in (x; y) = (x0; y0). We remark that this expression shows that rf(x0; y0) gives a normal vector to the level curve in the plane, i.e., rf 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 = rF (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 rF (x0; y0; z0) gives a normal vector to the level surface in the space, i.e., rF 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 rf = h2x; 2yi, we have rf(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. p p iii. We parametrizep thep 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 rf = h2xey; x2eyi, we have rf(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 rf = hy2 − 9x2y5; 2xy − 15x3y4i, we have rf(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 rF = h2x; 2y; 2zi, we have rF (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 rF = hyz − 2x; xz + z; xy + yi, we have rF (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 rF = h3x2; −2 sin z; −2y cos zi, we have rF (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).