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Math Help - Differentiation Applications with Cones Volume

  1. #1
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    Differentiation Applications with Cones Volume

    A small right cone of radius r cm and height h cm inscribed in a larger right cone of radius 6 cm and height 12 cm. The 2 bases are parallel and the vertex of the smaller cone lies at the centre of the larger cone base. The volume of the smaller cone is Vcm cube.

    (a) (i) Show that V = 4pi(r)sq - (2/3)pi(r)cube
    (ii) Determine the value of r and of h such that the smaller cone has the largest possible volume.

    (b) Given that the smaller cone has radius 3 cm and height 6 cm and water flows from a hole at the vertex of the larger cone to the smaller cone at a rate of pi/16 cm cube per sec. The height of the water in the smaller cone is k cm at time t seconds.

    (i) Show that the volume of the water in the smaller cone Vk , is (1/12)pi (k)cube at time t seconds
    (ii) Hence calculate the rate of change of k when k = 2.
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  2. #2
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    ***Draw a picture***

    (a) In general, V=\frac\pi{3}hr^2

    (i) By construction, the distance from the small cone's base to the large cone's apex is 12-h. The proportion between this length and the small cone's radius is the same as the proportion of height to radius in the large cone: \frac{r}{12-h}=\frac6{12} \to h=12-2r. Thus, V=\frac\pi{3}(12-2r)r^2=4\pi r^2-\frac23\pi r^3

    (ii) \frac{dV}{dr}=8\pi r - 2\pi r^2 = 0 \to r=h=4

    (b) In this cone, the radius:height ratio is 3:6, so \frac{r(k)}{k}=\frac12

    (i) V(k)=\frac\pi{3}k\left(\frac{k}2\right)^2=\frac\pi  {12}k^3

    (ii) V=\frac\pi{16}t=\frac\pi{12}k^3 Differentiating implicitly with respect to t, \frac{\pi}{16}=\frac{3\pi}{12}k^2\frac{dk}{dt} \to k^2k'=\frac14. So, when k=2, k'=\frac1{16}
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