# A plane flying horizontally at an altitude of 1 mi and a speed of 540 mi/h passes directly over a radar station. How do you find the rate at which the distance from the plane to the station is increasing when it is 5 mi away from the station?

Rate

Let us set up the following variables:

# {(s, "Horizontal distance of plane from the radar station (mi)"), (x, "Actual direct distance of plane from the radar station (mi)") :} #

Then, our aim is to find

The plane is moving in the horizontal direction at constant speed (

By Pythagoras;

# \ \ \ \ \ x^2 = s^2+1^2 #

# :.x^2 = s^2+1 # ..... [1]Differentiating Implicitly wrt

#t# we get:

# 2xdx/dt = 2s(ds)/dt + 0 #

# \ \ xdx/dt = s(ds)/dt #

# \ \ xdx/dt = 540s #

# \ \ xdx/dt = 540sqrt(x^2-1) " "# (Using [1])When

# x=5 => #

# \ \ \ \ \ 5dx/dt = 540sqrt(25-1) #

# :. 5dx/dt = 540sqrt(24) #

# :. dx/dt = 108sqrt(24) #

# :. dx/dt ~~ 529.1 # mi/hour

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To find the rate at which the distance from the plane to the station is increasing when it is 5 mi away from the station, we can use the concept of related rates.

Let ( d ) represent the distance between the plane and the radar station at any given time. Let ( t ) represent the time elapsed since the plane passed directly over the radar station.

Given: ( \frac{d}{dt} = 540 ) mph (speed of the plane) ( d = 1 ) mi (altitude of the plane) We need to find ( \frac{dd}{dt} ) when ( d = 5 ) mi.

Using the Pythagorean theorem, we have: [ d^2 = 1^2 + (540t)^2 ]

Differentiating both sides with respect to time ( t ), we get: [ 2d \frac{dd}{dt} = 2(540t)(540) ]

Substituting ( d = 5 ) mi, we can solve for ( \frac{dd}{dt} ): [ 2(5) \frac{dd}{dt} = 2(540t)(540) ] [ 10 \frac{dd}{dt} = 540^2t ] [ \frac{dd}{dt} = \frac{540^2t}{10} ] [ \frac{dd}{dt} = 291600t ]

When the plane is 5 miles away from the radar station, substitute ( d = 5 ) into the equation to find ( t ): [ 5^2 = 1^2 + (540t)^2 ] [ 25 = 1 + 291600t^2 ] [ 24 = 291600t^2 ] [ t^2 = \frac{24}{291600} ] [ t = \sqrt{\frac{24}{291600}} ]

Substitute the value of ( t ) back into the equation ( \frac{dd}{dt} = 291600t ) to find ( \frac{dd}{dt} ) when ( d = 5 ) mi.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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