Porisma

Review practice · Section 2 of 2

Questions 7–12 in this collection

Identify the parabola touched by every line in a family · Convert units before differentiating a conical pile · +4 more

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  1. 7

    The line y=mx+3/my=mx+3/m is tangent to the parabola y2=kxy^2=kx for every m≠0m\ne0. Find kk and the coordinates of contact in terms of mm. marks

    [2014/VJC/I/1]

    CJC · Review problems · Question 7

  2. 8

    Sand from a chute forms a conical pile whose altitude is always 4/34/3 of its base radius.

    (i) How fast is its volume increasing when the radius is 11 m and increases at 1/8 cm s−11/8\text{ cm s}^{-1}? marks

    (ii) How fast is its radius increasing when it is 22 m and the volume increases at 1000 cm3 s−11000\text{ cm}^3\text{ s}^{-1}? marks

    CJC · Review problems · Question 8

  3. 9

    Point Q(x,y)Q(x,y) lies on 4x2+xy+y2=364x^2+xy+y^2=36, with y≥0y\ge0. The curve meets the x-axis at P,RP,R as drawn.

    (i) Show that triangle PQRPQR has area A=3yA=3y. marks

    (ii) Find dy/dxdy/dx in terms of x,yx,y. marks

    (iii) Hence find the x-value where AA is stationary and use the second derivative test to determine its nature. marks

    If xx increases at 88 units/s, find the rate of change of AA when x=0x=0. marks

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    Upper part, y≥0, of the closed curve 4x²+xy+y²=36, with x-axis intersections P and R and a general point Q(x,y) on the upper-left arc. No numerical intercepts are supplied in this given diagram.
    Upper part, y≥0, of the closed curve 4x²+xy+y²=36, with x-axis intersections P and R and a general point Q(x,y) on the upper-left arc. No numerical intercepts are supplied in this given diagram.
    Upper part, y≥0, of the closed curve 4x²+xy+y²=36, with x-axis intersections P and R and a general point Q(x,y) on the upper-left arc. No numerical intercepts are supplied in this given diagram.

    [2013/DHS/I/11]

    CJC · Review problems · Question 9

  4. 10

    A flower-bed for a large garden consists of an xx m by yy m rectangle topped by a semicircle of diameter xx m. A low wall takes 33 hours per metre along the straight parts and 99 hours per metre around the semicircular arc. A total of 180180 hours is available.

    Use differentiation to find x,yx,y giving maximum area. marks

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    Flower-bed made of a rectangle of width x and height y with a semicircle of diameter x on top. Only the bottom x, two sides y, and outer semicircular arc form the wall; the joining diameter is dashed and internal.
    Flower-bed made of a rectangle of width x and height y with a semicircle of diameter x on top. Only the bottom x, two sides y, and outer semicircular arc form the wall; the joining diameter is dashed and internal.
    Flower-bed made of a rectangle of width x and height y with a semicircle of diameter x on top. Only the bottom x, two sides y, and outer semicircular arc form the wall; the joining diameter is dashed and internal.

    [2008/I/7]

    CJC · Review problems · Question 10

  5. 11

    A decorative monument for the 2010 Youth Olympic Games consists of a complete globe on a closed cylindrical base of radius rr and height hh. The globe radius is 1.25r1.25r. To minimise cost, the total surface area is fixed at 36.5 m236.5\text{ m}^2.

    (i) Show that h=(146−33πr2)/(8πr)h=(146-33\pi r^2)/(8\pi r). marks

    Hence find exactly:

    (ii) the cylinder radius giving greatest total volume. marks

    (iii) that maximum volume. marks

    A sphere of radius RR has area 4πR24\pi R^2 and volume 4πR3/34\pi R^3/3.

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    Complete globe above a closed cylindrical base. The globe radius is 1.25r, cylinder radius r and height h. The base has both circular ends; the diagram indicates the full sphere, not a hemisphere.
    Complete globe above a closed cylindrical base. The globe radius is 1.25r, cylinder radius r and height h. The base has both circular ends; the diagram indicates the full sphere, not a hemisphere.
    Complete globe above a closed cylindrical base. The globe radius is 1.25r, cylinder radius r and height h. The base has both circular ends; the diagram indicates the full sphere, not a hemisphere.

    [2009/CJC/Promo/12]

    CJC · Review problems · Question 11

  6. 12

    A square lawn ABCDABCD has side 2626 m and a circular pond of radius 1111 m centred at its centre PP. A rectangular flower crate QTCSQTCS has corner CC at the square corner and opposite corner QQ on the pond. Let RR be the midpoint of DCDC on the downward centre line. The angle RPQ=θRPQ=\theta is measured from PRPR, with 0≤θ≤π/40\le\theta\le\pi/4. Its base area is X m2X\text{ m}^2.

    (i) Express XX in terms of θ\theta and show dX/dθ=11(sin⁡θ−cos⁡θ)(13−11sin⁡θ−11cos⁡θ)dX/d\theta=11(\sin\theta-\cos\theta)(13-11\sin\theta-11\cos\theta). marks

    (ii) For stationary values, show the angles satisfy tan⁡θ1=1\tan\theta_1=1 and sin⁡(θ2+α)=k\sin(\theta_2+\alpha)=k, where k,αk,\alpha are exact constants. Find θ1,θ2\theta_1,\theta_2. marks

    (iii) Classify both stationary values and find their areas. marks

    (iv) Four identical crates are placed at A,B,C,DA,B,C,D, with the rest outside the pond covered by grass. Find the smallest grass area to three significant figures. marks

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    Square ABCD of side 26 m with central circular pond of radius 11 m and centre P. R is the midpoint of bottom side DC. Q lies on the southeast arc, with angle RPQ=theta between downward PR and PQ. Rectangle QTCS occupies the bottom-right corner: T is on BC, S on DC and C is the square corner. Its area is X square metres. The angle is between zero and pi/4.
    Square ABCD of side 26 m with central circular pond of radius 11 m and centre P. R is the midpoint of bottom side DC. Q lies on the southeast arc, with angle RPQ=theta between downward PR and PQ. Rectangle QTCS occupies the bottom-right corner: T is on BC, S on DC and C is the square corner. Its area is X square metres. The angle is between zero and pi/4.
    Square ABCD of side 26 m with central circular pond of radius 11 m and centre P. R is the midpoint of bottom side DC. Q lies on the southeast arc, with angle RPQ=theta between downward PR and PQ. Rectangle QTCS occupies the bottom-right corner: T is on BC, S on DC and C is the square corner. Its area is X square metres. The angle is between zero and pi/4.

    [2022/SAJC/I/10]

    CJC · Review problems · Question 12

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