1. no trace exists correct. 2. hyperbola : z 2 y 2 = ellipse : y z2 = ellipse : 5. circle : y 2 +z 2 = 2

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1 grandi (rg38778) Homework 5 grandi () This print-out should have 3 questions. Multiple-choice questions may continue on the next column or page find all choices before answering.. points Classify the quadric surface z. points Find and identify the trace of the quadric surface x y z = 3 in the plane x =.. no trace exists correct. hyperbola : z y = 3. ellipse : y + z = y 4. ellipse : y +z = 5. circle : y +z = 6. hyperbola : y z =. ellipsoid. hyperbolic paraboloid 3. one-sheeted hyperboloid 4. two-sheeted hyperboloid 5. elliptic paraboloid correct 6. elliptic cone The trace on any horizontal plane on or above the xy-plane, i.e.,any plane z = c, c, is always a circle, the trace on any plane parallel to the coordinate planes x = andy = isalwaysaparabola. Consequently the quadric surface is an elliptic paraboloid. keywords: quadric surface, ellipsoid, elliptic paraboloid, hyperbolic paraboloid, onesheeted hyperboloid, two-sheeted hyperboloid, x 7. hyperbola : y z = 8. circle : y +z = To determine the trace of x y z = 3 in the plane x =, we set x = in obtaining x y z = 3, y z =. Since no (real) values of y and z satisy this equation, no trace exists. 3. points Find an equation for the surface consisting of all points P(x, y, z) equidistant from the point P(,, ) and the plane x = 3.

2 grandi (rg38778) Homework 5 grandi (). x +y 4x+6z 5 =. y +z +6x 4y 5 = 3. x +z 6y +4z 5 = 4. y +z 6x+4y 5 = correct 5. x +y +4x 6z 5 = 6. x +z +6y 4z 5 = The distance from P(x, y, z) to P(,, ) is x +(y +) +z, the distance from P(x, y, z) to the plane x = 3 is x + 3. P lies on the graph of x+3 = x +(y +) +z. After squaring and expanding both sides this becomes x +6x+9 = x +y +4y +4+z. an equation for the surface is y +z 6x+4y 5 =. Which one of the following vector functions has this space curve as its graph?. r(t) = cost, sint, sin4t. r(t) = tcost, tsint, t 3. r(t) = tcost, t, tsint 4. r(t) = tcost, tsint, t, t >, correct 5. r(t) = sin4t, cost, sint 6. r(t) = tcost, t, tsint, t >, 7. r(t) = sint, cost, sint 8. r(t) = cost, sint, sint The surface is a cone such that horizontal cross-sections intersect the surface in a circle. Ontheotherhand,thespacecurveexistsonly for z >. But of the given vector functions, only r(t) = tcost, tsint, t, t >, satisfies the equations z(t) = x(t) +y(t), z(t) >. 4. points A space curve is shown in black on the surface z keywords: surface, space curve, vector function, 3D graph, circular cylinder, 5. points Find a parametrization r(t) = x(t), y(t), z(t) of the straight line passing through the origin in 3-space whose projection on the yz-plane is a line with slope, its projection on the zx-plane is a line with slope 3, i.e., x y z y =, x z = 3.

3 grandi (rg38778) Homework 5 grandi () 3. r(t) = 3t, t, 6t. r(t) = t, 6t, 3t 3. r(t) = t, t, 6t 4. r(t) = 6t, 3t, t 5. r(t) = t, 6t, t 6. r(t) = 6t, t, t correct Writing we see first that r(t) = x(t), y(t), z(t), x(t ) = y(t ) = z(t ) = for some choice of parameter t because the line passes through the origin. On the other hand, the projection on the yz- and zx-planes are given by, y(t), z(t), x(t),, z(t), respectively. But the slope conditions z y =, x z = 3 ensure that these projections are straight lines so z = y, x = 3z, z(t) = y(t), x(t) = 3z(t). r(t) = 6t, t, t is a parametrization of the straight line- there could be many others, of course. keywords: space curve, vector function, straight line, line, projection, coordinate plane, 6. points Find a parametrization for the circle having radius 3 and center (, 5, 4) that lies in a plane parallel to the xy-plane.. r(t) = +3sint, 5 3cost, 4. r(t) = 3cost, 5, 4 3sint 3. r(t) = 3sint, 5 + 3cost, 4 correct 4. r(t) =, 5+3cost, 4 3sint 5. r(t) =, 5 3cost, 4+3sint 6. r(t) = +3cost, 5, 4+3sint If the vector function r(t) = x(t), y(t), z(t) tracesoutthecirclehavingradius3andcenter (, 5, 4) that lies in a plane parallel to the xy-plane, then z(t) = 4 because the plane containing (, 5, 4) and parallel to the xyplane has the equation z = 4. On the other hand, the projection x(t), y(t), of this circle on the xy-plane has the equation (x ) +(y +5) = 9, as a circle in the xy-plane since it will be centered at (, 5) and have radius 3. and so (x(t) ) +(y(t)+5) = 9, r(t) = 3sint, 5+3cost, 4 is one parametrization of the circle - there are others, of course.

4 grandi (rg38778) Homework 5 grandi () 4 keywords: vector function, space curve, circle, plane, radius, center circle, Find r(t) if 7. points r (t) = sinti costj+tk and r() = i+j+k.. ( cost+)i (sint )j+ ( 6t + ) k. costi+(sint+)j+ ( 6t + ) k 3. ( cost+)i+(sint+)j+ ( 6t + ) k 4. costi (sint )j+ ( 6t + ) k 5. ( cost + )i (sint )j + ( 6t + ) k correct We start by finding the antiderivative of, we have ( r (t)dt = r (t) = sinti costj+tk. ) ( (sin t)dt i ) ( cos tdt j+ = ( cost)i (sint)j+(6t )k+c where C is a vector constant of integration. However, since the antiderivative evaluated at t = is i + C and we are given that r() = i+j+k, we know that i+c = i+j+k. Therefore, C = i+j+k. So we have that r(t) = ( cost)i (sint)j+(6t )k+i+j+k = ( cost+)i (sint )j+ ( 6t + ) k. Find the derivative of when and f(t) = u(t) v(t) u(t) = i 3t j+4t 3 k v(t) = ti+costj+sintk.. f (t) = 6tcost+5t sint+4t 3 cost correct. f (t) = 6tcost+5t sint+4t 3 cost 3. f (t) = 3tcost+7t sint+4t 3 cost 4. f (t) = 3tcost+t sint+4t 3 cost 5. f (t) = 6tcost+7t sint+4t 3 cost The product rule for derivatives of dot products says that f (t) = u(t) v (t)+u (t) v(t). Now) for the given u(t) and v(t), tdt k u (t) = 6tj+t k and so v (t) = i sintj+costk. f (t) = (+3t sint+4t 3 cost) +( 6tcost+t sint), f (t) = 6tcost+5t sint+4t 3 cost. keywords: 8. points keywords: 9. points

5 grandi (rg38778) Homework 5 grandi () 5 Which one of the following integrals gives the length of the curve. c(t) = t i+tj, t t +dt correct 6t + dt 6t +dt 6t + dt 6t + dt 6t +dt Thearclengthforthecurvec(t),a t b, is given by the integral expression But when b a c (t) dt. c(t) = t i+tj. arc length = 6 3. arc length = 8 4. arc length = 4 5. arc length = The length of the curve between c(t ) and c(t ) is given by the integral Now when L = t t c (t) dt. c(t) = (sin4t)i+3tj+(cos4t)k, we see that c (t) = (4cos4t)i+3j (4sin4t)bk. But then by the Pythagorean identity, c (t) = (6+9) / = 5. L = 5dt = [ ] 5t. we see that c (t) = 4ti+bj. arc length = L =. the curve has arc length = 4 6t +dt.. points When C is parametrized by c(t) = (sin4t)i+3tj+(cos4t)k, find its arc length between c() and c().. arc length = correct. points Find the unit tangent vector T(t) to the graph of the vector function r(t) = 3sint, +4t, +3cost.. T(t) = 3cost, 4, 3sint. T(t) = 3 5 cost, 4 5, 3 5 sint 3. T(t) = 3sint, 4, 3cost

6 grandi (rg38778) Homework 5 grandi () 6 4. T(t) = 3 5 cost, 4 5, 3 5 sint correct 5. κ(t) = 8 correct (4t 3/ +6) 5. T(t) = 3sint, 4, 3cost 6. T(t) = 3 5 sint, 4 5, 3 5 cost TheunittangentvectorT(t)tor(t)isgiven by T(t) = r (t) r (t). Now when we see that r(t) = keywords: r(t) = 3sint, +4t, +3cost, r (t) = 3cost, +4, 3sint T(t) = 3 (cos t+sin t)+(4) = cost, 4 5, 3 5 sint. points Determine the curvature, κ, of the curve. κ(t) = r(t) = t i+4tk. 8t 4t +6. κ(t) = ( 4t +6 ) 3/. The curvature κ of the curve is given by Now in this case, and Since r(t) = t i+4tk κ(t) = r (t) r (t) r (t) 3. r (t) = ti+4k, r (t) = i. it follows that r (t) r (t) = 8j r (t) r (t) = 8 = 8. r (t) = (4t +6) /, κ(t) = 8 (4t +6) 3/. 3. points Find the curvature, κ(t), of the curve parametrized by r(t) = ti+3sintj+3costk. κ(t) = 3. κ(t) = κ(t) = 3 3. κ(t) = 4. κ(t) = 8 (4t +6) / 8t (4t +6) 3/ 4. κ(t) = 3 3 correct 5. κ(t) = 3

7 grandi (rg38778) Homework 5 grandi () 7 Now But In this case and r (t) = i+3costj 3sintk r (t) = 3sintj 3costk. κ(t) = r (t) r (t) r (t) 3. r (t) r (t) = 9i+6costj 6sintk r (t) r (t) = Since it follows that keywords: 9 +6 (cos t+sin t) = 7 = 3 3. r (t) = 4+9 = 3, κ(t) = 3 3 (3) (3/) = points For what value of x does the graph of f(x) = 4 ex have maximum curvature?. x =. x = 4 ( 3. x = ln ) correct ( 6. x = ln 4 ) The curvature of the graph of y = f(x) is given by κ(x) = But when f(x) = 4 ex, f (x) (+ f (x) ) 3/. f (x) = f (x) = 4 ex. κ(x) = 6e x (6+e x ) 3/. Now the maximum value of κ(x) occurs at a critical value of κ. By the Quotient Rule, however, κ (x) ( e x (6+e x ) 3/ 3e 3x (6+e x ) / ) = 6 (6+e x ) 3 = 6e x{ 6 e x (6+e x ) 5/ }, so the only critical value of κ occurs when e x = 6, i.e., when ( x = ln ) Since the graph of e x becomes flatter as x and more vertical as x, the graph of y = f(x) will have maximum curvature at this critical value. 5. points Findequationsofthenormalplanetox = t, y = t, z = t 3 at the point (,4,8).. x+4y +z 4 = correct. 4. x = 4. x+y 4z 4 = 5. x = ln4 3. x+4y +z =

8 grandi (rg38778) Homework 5 grandi () 8 4. x+4y +z +4 = 5. x 4y +z 4 = 6. points At what point on the curve x = t 3, y = t, z = t 4 is the normal plane parallel to the plane 3x+y 4z = 9?. (,,) correct Letu = sinx. Thendu = cosxdx, In this case x = = u =, x = π 6 = u =. 3 [ ] +u du = 3 tan u. 3 4 π.. (,,) 3. (,,) 8. points Evaluate the definite integral 4. (,,) 5. (,, ). 4 3 π/ (cosθ +cos 3 θ)dθ. 7. points. Evaluate the integral π/6 6cosx +4sin x dx π correct correct π Since cos θ = sin θ the integral can be rewritten as π cosθ +cos 3 θ = cosθ(+cos θ) = cosθ(+ sin θ) = cosθ(3 sin θ).

9 grandi (rg38778) Homework 5 grandi () 9 π/ Now set x = sinθ. For then cosθ(3 sin θ)dθ. in which case π/4 π/4 [ e e u u du = ] π/4 π/4. In this case, dx = cosθdθ, θ = = x =, θ = π (3 x )dx = = x = points Evaluate the integral. eπ/ + e π/. eπ/ e π/ 3. 3 eπ/ + 3 e π/ 4. 3 eπ/ + 3 e π/ 5. 3 eπ/ 3 e π/ [3x 3 x3] e arctany +y dy. 6. eπ/ e π/ correct Set u = arctany. Then du = +y dy,. (eπ/ e π/ ).. points Evaluate the integral ln ln3 x + x dx ln correct 4. 7 ln 5. 5 ln ln 3 After division In this case = x + x = (x )+3 x 3 = x x + 3 x = x++ 3 x. ( x++ 3 x ) dx [ x +x+3ln x ] 3 = ( ) 5 ( ) 4 +3 ln ln.

10 grandi (rg38778) Homework 5 grandi () 7 +3ln. [u ] π/4 sinu.. points Evaluate the definite integral. π 4 8. π 3. π 4 4x x dx. π.. points Evaluate the integral.. ln 4 3 ln 5 ln5 ln e x dx. e x 3. ln π π 4 6. π correct Set x = sinu. Then dx = cosudu, In this case x = cosu, x = = u =, x = = u = π 4. π/4 = 4 4sin ucosu cosu π/4 = sin udu π/4 du ( ) cosu du. 4. ln 5 5. ln correct 6. ln Set u = e x. Then du = e x dx, so In this case 5 dx = e x du = u du, x = ln = u =, x = ln5 = u = 5. ( u u /u ) du = 5 u du. To evaluate this last integral we use partial fractions. For then u = ( u ). u+

11 5 grandi (rg38778) Homework 5 grandi () ( u ) du = [ ln u u+ u+ { ln 3 ln } 3 = ln. ] points Determine the integral sin x dx. x. xsin x x+c. xsin x+ 3. xsin x+ 4. xsin x 4 x +C x +C x +C 5. xsin x+ 4 x+c 6. xsin x+ x+c correct Set u = x. Then in which case du = x dx, sin udu. We now integrate by parts: usin u u du u = usin u+ u +C. xsin x+ x+c with C an arbitrary constant.

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