MATH113 Calculus and Geometry

Test 2016: Solutions

  • 1.

    i) 𝐚𝐛=(1,0,2)(-2,2,1)=0. [2]

    𝐚×𝐛=(1,0,2)×(-2,2,1)=(-4,-5,2). [2]

    ii) The point C is OA+OB=(2,1,0)+(3,1,1)=(5,2,1). [2]

    iii) The area of OACB is |OA×OB|. [1] We have OA×OB=(1,-2,-1). [2] The length of this vector is 6. [1]

    iv) We have cosθ=𝐮𝐯|𝐮||𝐯|=3|𝐮||𝐯| [1] and |𝐮×𝐯|=|𝐮||𝐯|sinθ. [1]

    Further, |𝐮×𝐯|=|(1,2,2)|=3. [1]

    Thus, cosθ=3sinθ|𝐮×𝐯|=33sinθ, and hence tanθ=33=3. [1]


  • 2.

    i) To calculate the arc length, we recall the formula: L=03|γ(t)|𝑑t. [1]

    Here γ(t)=(et-e-t,-2) [1], hence |γ(t)|=(et-e-t)2+4=e2t+e-2t+2. [1] This equals et+e-t. [1] It follows that the arc length is 03(et+e-t)𝑑t=e3-1e3. [1]

    ii) We have γ(0)=(2,5) [1] and γ(0)=(0,-2) [1]. So an equation of the tangent line to the image of γ at the point γ(0) is (x,y)=γ(0)+λγ(0)=(2,5)+λ(0,-2). [1]


  • 3.

    The surface is f(x,y,z)=0, where f(x,y,z)=x4-y2z2. [1]

    We have f=(4x3,-2yz2,-2zy2) [1] and f(2,2,2)=(32,-16,-16). [1]

    So the normal line is {(2,2,2)+λ(2,-1,-1)|λ} [1]

    and the tangent plane is (2,-1,-1)(x,y,z)=(2,-1,-1)(2,2,2)=0. [2]


  • 4.

    i) We check the partial derivatives ([1] for knowing the right condition). We have: (f2)z=2z and (f3)y=2y [1], so the partial derivatives do not match and hence 𝐟 cannot be expressed as ϕ. [1]

    ii) Here (g1)y=2x-z, (g2)x=2x-z. [1] Similarly (g1)z=-y-4z, (g3)x=-4z-y. [1] Finally, (g2)z=6y-x, (g3)y=6y-x [1], so all of the conditions are satisfied and hence 𝐠 can be expressed as ϕ for some ϕ. To find ϕ, since ϕx=2xy-2z2-yz, we obtain: ϕ=x2y-2xz2-xyz+h(y,z). [1] Now, checking against ϕy=g2, ϕz=g3, we obtain:

    ϕy=x2-xz+hy=x2+6yz-xz,ϕz=-4xz-xy+hz=3y2-4xz-xy[𝟏]

    Hence we have: hy=6yz and hz=3y2, so we deduce: h(y,z)=3y2z (+c). It follows that ϕ(x,y,z)=x2y-2xz2+3y2z-xyz. [1]


  • 5.

    Let f(x,y)=x2+2y2. We form the auxiliary function Λ(x,y,λ)=x2+2y2-λ(x2+x2-1). [2] Then we look for stationary points:

    Λx=2x-2λx=0,Λy=4y-2λy=0[𝟐]

    By the first equation, we have x=0 or λ=1. [1]

    If x=0, then y2=1 (since Λλ=0) and hence y=±1.

    If λ=1, then y=0 (by the second equation) and hence x=±1.

    So the possible extreme points are (0,1),(0,-1),(1,0),(-1,0). [1]

    We have f(0,1)=2, f(0,-1)=2, f(1,0)=1, f(-1,0)=1.

    We deduce that the greatest value of f is 2 [1], and the least value is 1. [1]


  • 6.

    The region D is given by: 0r3, 0θπ/2 [2], so

    I=0π/2031r2+1rdrdθ.  [𝟐]

    Now

    03rr2+1𝑑r=[12log(r2+1)]03=12log10,[𝟏]

    so   I=π/4log10. [1]