allie's car travels east at 40mi/hr while nick's car travels north at 30mi/hr from the same starting point. how fast is the distance between the two cars changing 10 minutes later?
distance D(t) = √((40t)² + (30t)²)
= √(1600t² + 900t²)
= √(2500t²)
= 50t
D'(t) = 50
When t = ⅙ hour, distance between cars is increasing by 50 mph
dx/dt = 40 mph
dy/dt = 30 mph
10 min = 10/60 = 1/6 hour
In 10 minutes , Allie would have traveled 40*1/6 = 6.67 miles and Nick would have traveled 30*1/6 = 5 miles
x= 6.67 and y=5
z = sqrt(x^2+y^2)
z = sqrt( (6.67)^2 + (5)^2 ) = 8.336
z^2 = x^2 + y^2
2z dz/dt = 2x dx/dt + 2 y dy/dt
z dz/dt = x dx/dt + y dy/dt
(8.336) dz/dt = (6.67) (40) + (5)(30)
dz/dt = ( (6.67) (40) + (5)(30)) / 8.336
= 50 mi/hr
https://gyazo.com/9a762653550645eb95bb01226b6d8960
Comments
distance D(t) = √((40t)² + (30t)²)
= √(1600t² + 900t²)
= √(2500t²)
= 50t
D'(t) = 50
When t = ⅙ hour, distance between cars is increasing by 50 mph
dx/dt = 40 mph
dy/dt = 30 mph
10 min = 10/60 = 1/6 hour
In 10 minutes , Allie would have traveled 40*1/6 = 6.67 miles and Nick would have traveled 30*1/6 = 5 miles
x= 6.67 and y=5
z = sqrt(x^2+y^2)
z = sqrt( (6.67)^2 + (5)^2 ) = 8.336
z^2 = x^2 + y^2
2z dz/dt = 2x dx/dt + 2 y dy/dt
z dz/dt = x dx/dt + y dy/dt
(8.336) dz/dt = (6.67) (40) + (5)(30)
dz/dt = ( (6.67) (40) + (5)(30)) / 8.336
= 50 mi/hr
https://gyazo.com/9a762653550645eb95bb01226b6d8960