Related Rates problem, Calculus?

Question

A plane flying horizontally at an altitude of 5 mi and a speed of 510 mi/h passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it is 10 mi away from the station. (Round to the nearest whole number.)

This question confused me a bit, like I know to draw a triangle and stuff, but A. I don't know how to use it, and B. When it says 10 miles away, does it mean horizontally, or 10 direct miles, because it's the difference between being the side or the hypoteneuse.

Thanks in advance :)

Faz · Accepted Answer

Note that the height (altitude) is constant so using Pythagarus:
x² + 5² = r²

Differentiate, w.r.t to t:
2x(dx/dt) = (2r)(dr/dt)
2(√75)(510) = 2(10)(dr/dt)  .................(Note that x=√75)
dr/dt = 255√3 = 441.67 mph

active open programming · Answer

Yes, I think that the question could have been more clear. I am assuming that the 10 miles away is the distance along the diagonal.

d = sqrt( x^2 + y^2 )

dd/dt = 1/2 * ( x^2 + y^2 )^( -1/2 ) * ( 2x * dx/dt + 2y * dy/dt )

d = 10, y = 5, dx/dt = 510, dy/dt = 0 (no vertical movement)

d = sqrt( x^2 + y^2 )
10 = sqrt( x^2 + 5^2 )
10^2 = x^2 + 5^2
100 = x^2 + 25
x = sqrt( 75 ) = 8.66

dd/dt = 1/2 * ( x^2 + y^2 )^( -1/2 ) * ( 2x * dx/dt + 2y * dy/dt )
dd/dt = 1/2 * ( 8.66^2 + 5^2 )^( -1/2 ) * ( 2(8.66) * 510 + 2(5) * 0 )
dd/dt = 1/2 * ( 8.66^2 + 5^2 )^( -1/2 ) * ( 2(8.66) * 510 )
dd/dt = ( 8.66^2 + 5^2 )^( -1/2 ) * ( (8.66) * 510 )
dd/dt = 441.67 mi/h approximately

Anonymous · Answer

[a] floor section = circumference * top = 2? r * h top = 10cm + 0.1cm * t h = 10 + 0.1t A = 10? * (10 + 0.1t) = 100? + ? t dA/dt = fee of exchange of section with relation to time dA/dt = (100? + ? t)' = ? cm^2/sec [b] quantity = portion of base * top = ? r^2 * (10 + 0.1t) V = 25? (10 + 0.1t) V = 250? + 2.5? t dV/dt = 2.5? cm^3/sec wish this permits!

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Related Rates problem, Calculus?

3 Answers