Midterm Review II Math , Fall 2018

Transcription

1 Midterm Review II Math , Fall 218 The test will cover section 12.5 of chapter 12 and section of chapter 13. Examples in class, quizzes and homework problems are the best practice for the test. All problems listed below are very similar to examples in class or homework problems. I will give the details for some problems and the hints (step-by-step with explanations) of the solution with details needed to be filled in. Notice: you do not have to give as much explanations as I do in the test. Section 12.5 Find parametric and symmetric equations of lines, equations of planes. We have some types of problems related to equations of lines: The key point for a line equation is to find A POINT and A DIRECTION VECTOR. The you can use the formulas to get the parametric or symmetric equation. 1. Find a parametric (or symmetric) equation of the line which passes through a point (1, 2, 3) and is parallel to the vector 4, 5, 6. (In this case, the direction vector of the line is 4, 5, 6 ). A direction vector of the line is a vector which is parallel to the line. Hence, 4, 5, 6 is the direction vector. A parametric equation of the line: x = 1 + 4t, y = 2 + 5t, z = 3 + 6t. The symmetric equation of the line: x 1 4 = y 2 5 = z Find a parametric (or symmetric) equation of the line (L) which passes through a point (1, 2, 3) and is parallel to the line (L ) x = 1 + t, y = 3t, z = 2 + 5t. L and L are parallel, so they can share same direction vectors. A direction vector of (L ) is,, which is also a direction vector of (L). 3. Find a parametric (or symmetric) equation of the line (L) which passes through two points P (1, 2, 3) and Q(4, 5, 6). A point is either P (1, 2, 3) or Q(4, 5, 6). Let us pick P (1, 2, 3).

2 Because the line passes through P and Q, so a direction vector can be chosen as the vector P Q =. 4. Find a parametric (or symmetric) equation of the line L which passes through a point (1, 2, 3) and is perpendicular to the plane x y + 2z 1 =. A point is P (1, 2, 3). The plane is perpendicular to its normal vectors and L is perpendicular to the plane. Hence, L and the plane s normal vectors are parallel. Then we can take the normal vector,, of the plane to be a direction vector. 5. Find a parametric (or symmetric) equation of the line which passes through a point (1, 2, 3) and is perpendicular to vectors 1,, 3 and 2, 1, 5. A point is P (1, 2, 3). The line L is perpendicular to 1,, 3 and 2, 1, 5, so L is parallel to the cross product 1,, 3 2, 1, 5. Hence, 1,, 3 2, 1, 5 =,, is a direction vector of L. 6. Find a parametric (or symmetric) equation of the line (L) which is the intersection of two planes (p 1 ) x + y + z 1 = and (p 2 ) 2x y z =. The line is the intersection of two planes, so a point which is in the line must be in the intersection of two planes. One way to get a point is to set z = and solve for x and y. Let z =. We have two equations and two variables x + y 1 = and 2x y =, so x = 1/3 and y = 2/3. A point on the line L is (1/3, 2/3, ). The line L is is in the plane (p 1 ), so the normal vector n 1 = 1, 1, 1 of (p 1 ) is perpendicular to L. Similarly, The line L is is also in the plane (p 2 ), so the normal vector n 2 = 2, 1, 1 of (p 2 ) is perpendicular to L. Hence, L is parallel to the cross product n 1 n 2. Then a direction vector of L is n 1 n 2 =,,. We have some types of problems related to equations of planes: The key point for a plane equation is to find A POINT and A NORMAL VECTOR. The you can use the formulas to get an equation of the plane.

3 1. Find an equation of the plane which passes through the point (1, 2, 3) and is perpendicular to the vector 4, 5, 6. A normal vector of a plane is a vector that is perpendicular to the plane. Hence, 4, 5, 6 is a normal vector of the plane. Write the equation of the plane: 4(x 1) + 5(y 2) + 6(z 3) =. Hence, 4x + 5y + 6z 32 =. 2. Find an equation of the plane which passes through the point (1, 2, 3) and is perpendicular to the line (L) x = 1 + t, y = 3t, z = 2 + 5t. The line L is perpendicular to the plane. Hence, a direction vector,, of (L) is also a normal vector of the plane. Write the equation of the plane: 3. Find a parametric (or symmetric) equation of the plane (p) which passes through a point (1, 2, 3) and is parallel to the plane (p ) x + y + z 1 =. Two parallel planes share same normal vectors. Hence, the normal vector,, of (p ) is also a normal vector of (p). 4. Find an equation of the plane (p) which passes through P (1, 3, 2), Q(3, 1, 6) and R(5, 2, ). A point is either P (1, 3, 2), Q(3, 1, 6) or R(5, 2, ). Let us pick P (1, 3, 2). A normal vector n of (p) is perpendicular to every vectors in (p), so n is parallel to the cross product of two arbitrary non-parallel vectors in (p). We pick n = P Q P R =,,,, =,,. 5. Find an equation of the plane (p) which passes through the point (1, 2, 3) and is parallel to two lines (L 1 ) x = 1 + t, y = 3t, z = 2 + 5t and (L 2 ) x = 1 + t, y = 2 t, z = 4 3t. The lines L 1 and L 2 are parallel to (p), so a normal vector n of (p) is perpendicular to the direction vector a 1 =,, of L 1 and a 2 =,, of L 2. Hence, the normal vector n = a 1 a 2 =,,.

4 6. Find an equation of the plane (p) which passes through the point P (3, 5, 1) and contains the line (L) x = 4 t, y = 1 + 2t, z = 3t. A point is P (3, 5, 1). We need to find two non-parallel vectors on the plane (p) to make a normal vector of (p). A very natural one is the direction vector of a =,, of L. To find another vector one the plane which is not parallel to a, we pick a point Q(,, ) on L with t =. Hence, P Q =,, is the other vector on (p). The normal vector n = a P Q =,,. 7. Find an equation of the plane (p) which contains the line (L) x 1 = y 3 2 = z and is perpendicular to the plane (p ) x + y 2z = 1. The plane (p) contains the line L. Any point on L is in (p). Set x 1 =, i.e, x = 1. Then y = 3 and z =. A point is (1, 3, ). Since (p) is perpendicular to (p ), the normal vector n =,, is parallel to (p). On the other hand, (p) contains L, so the direction vector a =,, of L is also parallel to (p). Hence, the normal vector of (p) n = n a =,,. Find the angle between two lines, two planes or between a plane and a line. 1. Find the angle between two planes (p 1 ) x + y + z = 1 and p 2 x 2y + 3z = 1. The angle of two planes is the angle of their normal vectors. The normal vector of (p 1 ) n 1 =,,. The normal vector of (p 2 ) n 2 =,,. Using the dot product formula to find the angle θ of n 1 and n 2, Hence, θ =. cos(θ) = n 1 n 2 n 1 n 2 =. Find the distance of a point to a plane, two parallel planes and two skew lines. 1. Find the distance from P (1, 2, 3) to the plane 2x + 5y + z = 1. Using the distance formula of a point P (x 1, y 1, z 1 ) to the plane ax + by + cz + d =, D = ax 1 + by 1 + cz 1 + d a2 + b 2 + c 2

5 The formula will be given for the test. For our problem, D = (2)(1) + (5)(2) + (1)(3) =. 2. Find the distance between two parallel planes (p 1 ) x + y + z = 5 and (p 2 ) 2x + 2y + 2z = 9. Since two planes are parallel, the distance of two planes and equal to the distance of a point on one plane to another one. We have 2 steps: Find a point on (p 1 ): Set x = y =, then z = 5. We have the point P (,, 5). Using the distance formula to find the distance from P to (p 2 ) as the previous problem, D = =. 3. Find the distance between two skew lines (L 1 ) x = 1+t, y = 2+3t, z = 4 t and (L 2 ) x = 2s, y = 3 + s, z = 3 + 4s. Step 1. Find an equation of the plane (p) which contains (L 1 ) and is parallel to (L 2 ): Any point on (L 1 ) is on (p). Let t =, the point P (1, 2, 4) (p). The direction vector of L 1 a 1 =,,. The direction vector of L 2 a 2 =,,. The normal vector of (p) n = a 1 a 2 =,,. Write the equation of the plane (p): Step 2. Find a point P on (L 2 ). Set s =, P (,, ). Step 3. Find the distance of P (,, ) to (p): D = =. Section 13.1 There may not be any problem which are directly from this section in the test, but you should understand vector functions, space curves and their related properties to do problems in section 13.2 and Examples in class and problems of homework 7 are a good source to practice. Section 13.2 Calculate tangents, unit tangents and tangent lines of vector functions and space curves. 1. For the curve (C) with respect to the vector function r(t) = (1+t 3 )i+te t j+ sin(2t)k,

6 (a) Find the tangent r (t) of (C) and the tangent line of (C) at the point where t =. (b) Find the unit tangent vector of (C) at general point and at the point where t =. The problem is similar to one in Quiz 5. See the solution of QUIZ For the curve (C) with respect to the vector function r(t) = 2 cos(t)i + sin(t)j + tk, find the tangent, the unit tangent vector and the tangent line at the point P (, 1, π/2). The problem is similar to one in Quiz 5. See the solution of QUIZ 5. Calculate integrals of vector functions. 1. Let r(t) = 2 cos 2 (t)i + sin 4 (t)j + t 2 + 1k, calculate r(t)dt. r(t)dt = ( 2 cos 2 (t)dt)i + ( sin 4 (t)dt)j + ( t2 + 1dt)k =. Section 13.3 Calculate the arc length of a space curve. 1. Let the curve (C) be with respect to the vector function r(t) = cos(t)i + sin(t)j + tk, calculate the arc length of C from (1,, ) to (1,, 2π). At (1,, ), t =. At (1,, 2π), t = 2π. The arc length L = ( sin(t))2 + (cos(t)) dt = 2dt = 2 2π. Reparametrize a vector function with respect to the arc length. 1. Reparametrize the helix r(t) = cos(t)i+sin(t)j+tk with respect to arc length measured from (1,, ) in the direction of increasing t. Step 1. At (1,, ), t =.

7 Step 2. Calculate the arc length function s. s(t) = t r (u) du = Hence, s = 2t. t Step 3. Solve for t. We have t = s/ 2. Step 4. Substitute t = s/ 2 into r(t). t ( sin(t))2 + (cos(t)) dt = 2dt = 2t. r(t(s)) = cos(s/ 2)i + sin(s/ 2)j + (s/ 2)k. Calculate the curvature using the formulas (will be given in the test) κ(t) = T (t) r (t) = r (t) r (t) r (t) Find the curvature of the vector function r(t) = 2ti + e t j + e t k. You can use any formula you like. Here I choose the more convenient one: κ(t) = r (t) r (t) r (t) 3. Step 1. Calculate r (t) = 2, e t, e t. Step 2. Calculate r (t) =. Step 3. Calculate r (t) =, e t, e t. Step 4. Calculate r (t) r (t) =. Step 5. Calculate r (t) r (t) =. Step 6. Calculate κ(t) =. 2. Find the curvature of the vector function of the twisted cubic r(t) = t, t 2, t 3 at a general point and at the point (,, ). it is similar to the previous problem for the κ(t). At the point (,, ), t =, we plug in t = into the κ(t). Calculate normal vectors, binormal vector, normal planes and osculating planes. 1. For the helix r(t) = cos(t)i + sin(t)j + tk, (a) Find the unit tangent vector T (t), the normal vector N(t) and the binormal vector B(t). EXCEPT the formula of the unit tangent vector, the others will be given in the test. Step 1. Calculate r (t) = sin(t), cos(t), 1. Step 2. Calculate r (t) = ( sin(t)) 2 + (cos(t)) = 2. Step 3. Calculate T (t) = r (t) r (t) = sin(t)/ 2, cos(t)/ 2, 1/ 2. Step 4. Calculate T (t) = cos(t)/ 2, sin(t)/ 2,. Step 5. Calculate T (t) = 1/ 2. Step 6. Calculate the normal vector N(t) = T (t) T (t) = cos(t), sin(t),.

8 Step 7. Calculate the binormal vector B(t) = T (t) N(t) =. (b) Find the equations of the normal plane and osculating plane at P (, 1, π/2). At P (, 1, π/2), t = π/2. A normal plane at the point P is the plane that consists all the lines through P and is perpendicular to the unit tangent vector T. Hence, The normal vector n = T (π/2) = 1/ 2,, 1/ 2 Write the equation: An osculating plane at the point P is the plane that consists all the lines through P and is perpendicular to the binormal vector B. Hence, The normal vector n = B(π/2) =. Write the equation: