1 .(6pts Find symmetric equations of the line L passing through the point (, 5, and perpendicular to the plane x + 3y z = 9. (a x = y = z (b x (c (x = ( 5(y 3 = z + (d x (e (x + 3(y 3 (z = 9 = y 3 5 = z + = y 3 5 = z + The symmetric equations of line are given by (x x 0 /a = (y y 0 /b = (z z 0 /c, where (x 0, y 0, z 0 is a point on the line and a, b, c is a direction vector. Since L is perpendicular to the plane x + 3y z = 9, then we can take the normal to the plane as the direction vector, this is,, 3, is a direction vector of L. Therefore, the symmetric equations are x = y + 5 = z 3. = 9.(6pts The two curves below intersect at the point (, 4, = r (0 = r (. Find the cosine of the angle of intersection ( r (t = e 3t i + 4 sin t + π j + (t k r (t = ti + 4j + (t k (a 5 (b 0 (c 5 (d e e + 4 (e 3 Note r (t = ( 3e 3t, 4 cos t + π, t r (t =, 0, t To compute the angle of intersection we find r (0 = 3, 0, 0 r ( =, 0, so that cos θ = r (0 r ( r (0 r ( = =. 5 3.(6pts Find the projection of the vector,, 5 onto the vector,, 4 (a,, 4 (b,, 5 (c 6, 3, (d 3, 3, 5 (e,, 5 5 proj,,4 (,, 5 =,, 4,, 5,, 4,, 4,, 4 =,, 4 =,, 4
2 4.(6pts Find Recall: r(xdx where sec x dx = tan x + C. r(x = (sec xi + e x k (a (tan x + C i + C j + (e x + C 3 k (b (tan x + C i + (e x + C k (c (tan xi + e x k (d tan x + e x + C (e (tan x + Ci + Cj + (e x + Ck ((sec r(xdx = xi + e x k dx ( ( = sec xdx i + ( 0dx j + e x dx k = (tan x + C i + C j + (e x + C 3 k 5.(6ptsThe curvature of the function y = sin x at x = π is (a (b 0 (c (d (e Does not exist. ( π Let r(t = t, sin t, 0. Then r (t =, cos t, 0, r (t = 0, sin t, 0, r = ( π, 0, 0, and r = 0,, 0. One of the formulas for curvature says that κ = r r r 3. But r r at t = π is 0 0 = 0, 0, and 0, 0, = (6pts What is the (approximate normal component of the acceleration for a car traveling 9 m/s around a curve with curvature κ = /3 m. (a 54.3 m/s (b.9 m/s (c m/s (d 400. m/s (e 40.3 m/s Using the formula that a N = κv. Plugging in the values from the problem we find that a N = (6pts Find the area of the triangle formed by the three points (, 0,, (, 0, and (3, 3, 3. (a 3 (b 0 (c 4 (d 3 (e.
3 Two vectors which form two sides of the triangle are, 0, =, 0,, 0, and, 3, = 3, 3, 3, 0,. Hence 0 = 3 0, (, 3 0 = 3, 0, 3 3 The area of the parallelogram is 3, 0, 3 = = 3 and the area of the triangle is half this. 8.(6pts Find the volume of the parallelepiped spanned by the three vectors,,, 0,, and 3,,. (a (b 3 (c 9 (d 3 (e 0 Answer is the absolute value of the triple product 0 3 = ( = = 9.(0pts Find an equation for the plane which goes through the point (,, 5 and contains the line,, t + 3, 4,. One vector in the plane is,,. A second is 3, 4,,, 5 =,, 4. Hence a normal vector for the plane is N = 4 = + 4, ( + 8, 4 =, 6, Hence an equation is, 6, x, y, z =, 6,,, 5 = 0 = 0 or x + 6y + z = 0 or x + 3y + z = 0. 0.(0pts Let Π be the plane passing through the point Q = (5, 0, 3 with normal vector n = 3,,. Let P be the point on Π which is closest to the origin. Let l be the line passing through the origin and P. Hint: parts (a and (b can be answered independently of each other. (a Find the co-ordinates of P. (b Find a vector equation for l. We find the position vector of P as OP. = Proj n OQ = OQ n n n n = 3,, = 6,, and so P = (6,,. The line l has direction n, and passes through the origin, so has vector equation 3,, x, y, z = 0
4 .(0pts Suppose the curve C has parametric equations: x(t = t 3 t, y(t = t, z(t = t + t (a Let P = (0,,. Find t 0 so that P = (x(t 0, y(t 0, z(t 0. (b Let r(t = t 3 t, t, t + t. Find r (t 0, the tangent vector to the above curve C at the point P = (0,,. (c Find the parametric equation for the tangent line to the above curve C at the point P = (0,,. (a 0 = t 3 0 t 0 = t 0 (t 0 (t 0 + implies t = 0,, or. = t 0 implies t 0 = i.e. t 0 =. (b r (t = 3t, t, t + r ( =,, 3 (c Let v(t =,, 3. Then the vector equation for the tangent line to C at P is given by tv( + 0,, = t, t, 3t + 0,, = t, t, 3t +. Then the parametric equation for the tangent line to C at P is given by x(t = t, y(t = t, z(t = 3t +.
5 .(0pts Suppose a particle has acceleration function a(t = j (sin tk for t 0. (t + Suppose that the initial position is r(0 = k and the initial velocity is v(0 = i + j + k. (a Find the velocity function. (b Find the position function. First find the velocity function as the integral of the acceleration function ( v(t = a(tdt = C i + t + + C j + (cos t + C 3 k Next we plug in t = 0 and compare with the initial velocity ( v(0 = C i C j + (cos 0 + C 3 k = C i + ( + C j + ( + C 3 k and by comparison to the given initial velocity, v(0 = i + j + k, we have C =, C = C 3 = 0 So, the velocity function is v(t = i + j + (cos tk t + The position function is the integral of the velocity function r(t = v(tdt = (t + D i + (ln(t + + D j + (sin t + D 3 k Next, we plug in t = 0 and compare with the initial position r(0 = (0 + D i + (ln(0 + + D j + (sin 0 + D 3 k = D i + D j + D 3 k and by comparison to the given initial position, r(0 = k, we have Thus, the position function is D = D = 0 D 3 = r(t = ti + ln(t + j + + sin tk OR using vector arithmetic v(t = 0, (t +, sin t dt = 0, Then v(0 = 0,, + C =,, so C =, 0, 0 and v(t =, t +, cos t t +, cos t + C Then r(t =, t +, cos t dt = t, ln t +, sin t + D Then r(0 = 0, 0, 0 + D = 0, 0, 0 so D = 0, 0, and r(t = t, ln t +, + sin t 3.(0pts Find an equation for the osculating plane of r(t = t, cos t, e t when t = 0. r (t =, sin t, e t, r (t = 0, cos t, e t, r(0 = 0,,, r (0 =, 0,, and r (0 = 0,,. The normal vector is, 0, 0,, =,,. So an equation is (x 0 (y (z = 0 or x y z =.
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