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(Solved): \[ v=\frac{1}{3} \pi h^{3} \sec ^{2} \theta \sin ^{2} \theta \] Where \( h \) is the height of the ...




\[
v=\frac{1}{3} \pi h^{3} \sec ^{2} \theta \sin ^{2} \theta
\]
Where \( h \) is the height of the cone h metres and \( \thet
\[ v=\frac{1}{3} \pi h^{3} \sec ^{2} \theta \sin ^{2} \theta \] Where \( h \) is the height of the cone h metres and \( \theta \) (in radians) is the angle shown in Figure 1. Note that \( V, h \) and \( \theta \) are all tunctions of time. (a) Show that \( \frac{d V}{d t}=\frac{1}{3} \pi h^{2}\left[2 h \tan \theta \sec ^{2} \theta \frac{d \theta}{d t}+3 \frac{d h}{d t} \tan ^{2} \theta\right] \) (b) Consider the instant when \( \theta=\frac{\pi}{6} \) radians and \( V=3 \pi \) cubic metres. (i) Find \( h \) at this instant. (ii) At this instant, \( \frac{d \theta}{d t}=\frac{\pi}{12} \) radians per second and the height of the con is decreasing at approximately \( 1.81 \) metres per second. Find \( \frac{d v}{d \tau} \).


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Given that when h is the height of the cone in meters and ? is the angle then volume of
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