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H H Chau
Questions and answers by
H H Chau
Answers (14)
At x=1 km dθ/dt=3 rev/min=6π rad/min dx/dt=(1^2+4^2)/4 * 6π=80.1 km/min=4806.6 kph
Concur.
See another question. x=2.075 miles t=6.200 hours
Concur. At x=12 ft dx/dt=(12^2+130^2)/130 * 22π=9062 ft/min
tan(θ)=x/1 cos(θ)=1/sqrt(x^2+1) sec^2(θ)=x^2+1 dθ/dt=3 rev/min=6π rad/min x=tan(θ) dx/dt=sec^2(θ) dθ/dt At x=0.5 km dx/dt=(0.5^2+1)*6π=23.6 km/min=1414 kph
dθ/dt=2 rev/min=4π rad/min dx/dt=8*(sec^2(π/5))*8π=154 miles/min=9216 mph
Concur. dx/dt=7*(4/3)*2π=176 miles/min=10555miles/hr
Note to above attempt: velocity in x-direction, not radial velocity.
θπ cos(π/6)=sqrt(3)/2 sec^2(π/6)=4/3 dθ/dt=4 rev/min = 8π rad/min tan(θ)=x/13 x=13 tan(θ) dx/dt=13 sec^2(θ) dθ/dt At θ=π/6 dx/dt=13*(4/3)*8π=435.6 miles/min = 26138 miles/hr
Um ... sec^2(Ø)=(x^2+0.5^2)/0.5^2
Oops! Third line should be: ec^2(Ø)=(x^2+0.5^2)/0.5^2
tan(Ø)=x/0.5 cos(Ø)=0.5/sqrt(x^2+0.5^5) sec^2(Ø)=(x^2+0.5^2) dØ/dt=4 * 2π = 8π rad/min x=0.5 tan(Ø) dx/dt=0.5 sec^2(Ø) dØ/dt dx/dt=(x^2+0.5^2)/0.5 * 8π when x=0.25 miles dx/dt=(0.25^2+0.5^2)/0.5 * 8π = 5π = 15.7 miles/min (note to about
distance of boat rowing = sqrt(5.5^2 + x^2) distance of running = 6.5 - x total travel time t = sqrt(5.5^2 + x^2)/1.2 + (6.5 - x)/3.4 dt/dx = x/(1.2*sqrt(5.5^2 + x^2) - 1/3.4 d^2t/dx^2 = +ve at any x x is at its minimum when dt/dx=0 x/(1.2*sqrt(5.5^2 +
Consider a right angled triangle, tan(A)=x/3 cos(A)=3/sqrt(x^2+9) sec^2(A)=(x^2+9)/3^2 dA/dt=4 rev/min = 2*pi/15 rad/s Also, x=3 tan(A) dx/dt=3 sec^2(A) dA/dt dx/dt=(x^2+9)/3 * 0.419 At x=1, dx/dt=1.396 km/s or 5027 km/hr. The above attempt confused
