I need help for this problem. Thanks. I really appreciate the help. How do I find theta?
The minute hand moves 12 times as fast as the hour hand, because the hour hand moves 1/12 of the area of the clock face while the minute hand moves the whole area of the clock face. At 4 o'clock, the hands are 120 degrees apart (one third of 360 degrees). The tips of the hands at this time are distance x apart where x is given by the cosine rule: x=sqrt(3^2+4^2-2*3*4cos120)=sqrt(25+12)=sqrt(37), because cos120=-1/2.
At time t minutes after 4, the angle between the tips of the hands decreases initially. For example, if t=15 minutes, the hour hand will move 1/4 the distance between 4 and 5, i.e., 30/4=7.5 degrees, while the minute will have moved 90 degrees. The angle between the hands becomes 120-90+7.5=37.5 degrees. For t minutes, then, the angle ("theta") changes to 120-360t/60+30t/60 degrees=120-6t+t/2=120-11t/2. (Theta can be found for any time t minutes after 4 o'clock from this formula. Also, t can be found when theta is zero, the hands being aligned: t=240/11.) So, x=sqrt(25-24cos(120-11t/2))=sqrt(25-24(cos120cos(11t/2)+sin120sin(11t/2)). If t is very small cos(11t/2) is close to 1 and sin(11t/2) is close to 11t/2.
The change in x is sqrt(37-66tsqrt(3))-sqrt(37)=sqrt(37)(sqrt(1-66tsqrt(3)/37)-1).
We can use the Binomial Theorem to evaluate sqrt(1-66tsqrt(3)/37)=(1-66tsqrt(3)/37)^(1/2). When t is very small we can ignore t^2 terms and higher powers of t, so we get 1-33tsqrt(3)/37 as an approximation. The change in x becomes sqrt(37)(1-33tsqrt(3)/37-1)=-33tsqrt(3/37). The instantaneous rate of change of x at 4 o'clock is found by dividing this expression by t=-33sqrt(3/37)=-9.3967"/minute=-0.1566"/sec. The minus sign indicates that the hands are closing at the rate of 0.1566 inches per second.
The solution can also be found applying formal calculus, where dx (the change in x) is related to the angle change (dø) and the angular rate of change dø/dt can be related to the rate of change of x, dx/dt. This involves differentiating the cosine expression. Because a time of 4 o'clock has been specified, the solution has been calculated from first principles, thus avoiding calculus.
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