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An airplane is flying at an elevation of 6 miles on a flight path that will take it directly over a radar tracking station. Let S represent the distance between the radar station and the plane. If S is decreasing at a rate of 400 mph when S is 10 miles, what is the velocity of the plane

Sagot :

Boffeemadrididn

Answer:

[tex]-500\ \text{mph}[/tex]

Explanation:

h = Height at which the plane is flying = 6 miles

S = Distance between plane and radar = 10 miles

[tex]\dfrac{dS}{dt}[/tex] = Rate at which S is decreasing = -400 mph

Distance between S and and the elevation of the plane

[tex]b=\sqrt{S^2-h^2}=\sqrt{10^2-6^2}\\\Rightarrow b=8[/tex]

From Pythagoras theorem we get

[tex]S^2=b^2+h^2[/tex]

Differentiating with respect to time we get

[tex]2S\dfrac{dS}{dt}=2b\dfrac{db}{dt}+2h\dfrac{dh}{dt}\\\Rightarrow S\dfrac{dS}{dt}=b\dfrac{db}{dt}+h\dfrac{dh}{dt}\\\Rightarrow 10\times -400=8\times \dfrac{db}{dt}+0\\\Rightarrow \dfrac{db}{dt}=\dfrac{10\times -400}{8}\\\Rightarrow \dfrac{db}{dt}=-500\ \text{mph}[/tex]

Velocity of the plane is [tex]-500\ \text{mph}[/tex].

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Nuhulawal20idn

If the distance between the radar station and the plane is decreasing at the given rate, the velocity of the plane is -500mph

Given the data in the question;

  • Elevation; [tex]x = 6miles[/tex]
  • Distance between the radar station and the plane; [tex]S = 10miles[/tex]
  • Since "S" is decreasing at a rate of 400 mph; [tex]\frac{ds}{dt} = -400mph[/tex]

As illustrated in the diagram below, we determine the value of "y"

Using Pythagorean theorem:

[tex]S^2 = x^2 + y^2[/tex]------------Let this be Equation 1

We substitute in our value

[tex]y = \sqrt{(10miles)^2- (6miles)^2} \\\\y = \sqrt{100mi^2- 36mi^2}\\\\y = \sqrt{64mi^2}\\\\y = 8miles[/tex]

Now, we determine velocity of the plane i.e the change in distance in horizontal direction ([tex]\frac{dy}{dt}[/tex])

Lets differentiate Equation 1 with respect to time t

[tex]2S\frac{ds}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt}[/tex]------ Let this be Equation 2

Since, the plane is not landing, [tex]\frac{dx}{dt} =0[/tex]

We substitute our values into Equation 2 and find [tex]\frac{dy}{dt}[/tex]

[tex][2*10mi*(-400mph) ] = [2*6mi*0] + [2*8mi * \frac{dy}{dt}]\\\\-8000m^2ph = 0 + 16miles * \frac{dy}{dt}\\\\\frac{dy}{dt} = \frac{-8000m^2ph}{16miles}\\\\\frac{dy}{dt} = -500mph[/tex]

Therefore, if the distance between the radar station and the plane is decreasing at the given rate, the velocity of the plane is -500mph

Learn more: https://brainly.com/question/18187424

View image Nuhulawal20
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