GROVE Ex 6.2 Q 32 and 33 (1 Viewer)
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Q 32. An observer sees a plane directly overhead, 2 km above the ground. The plane is moving horizontally at a constant speed of 500kmh. Find the rate at which the distance will be increasing between the observer and the plane when the distance between them is 5 km.
<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{dx}{dt} &= 500 km h^{-1}\\ \\ D &= (x^2 @plus;4)^{\frac{1}{2}}\\ \\ \frac{dD}{dx} &= \frac{x}{\sqrt{x^2 @plus;4}}\\ \\ \\ \frac{dD}{dt} &= \left(\frac{dD}{dx} \right)\left(\frac{dx}{dt} \right )\\ \\ &= (\frac{x}{\sqrt{x^2 @plus;4}})(500)\\ \\ \\ &= \left(\frac{500x}{\sqrt{x^2 @plus;4}} \right)\\ \\ \text{When D} &= 5, x =\sqrt{21}\\ \\ \therefore \frac{dD}{dt} = \frac{(500)(\sqrt{21})}{\sqrt{(\sqrt{21})^2 @plus;4}}\\ &= 458 kmhr" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{dx}{dt} &= 500 km h^{-1}\\ \\ D &= (x^2 +4)^{\frac{1}{2}}\\ \\ \frac{dD}{dx} &= \frac{x}{\sqrt{x^2 +4}}\\ \\ \\ \frac{dD}{dt} &= \left(\frac{dD}{dx} \right)\left(\frac{dx}{dt} \right )\\ \\ &= (\frac{x}{\sqrt{x^2 +4}})(500)\\ \\ \\ &= \left(\frac{500x}{\sqrt{x^2 +4}} \right)\\ \\ \text{When D} &= 5, x =\sqrt{21}\\ \\ \therefore \frac{dD}{dt} = \frac{(500)(\sqrt{21})}{\sqrt{(\sqrt{21})^2 +4}}\\ &= 458 kmhr" title="\frac{dx}{dt} &= 500 km h^{-1}\\ \\ D &= (x^2 +4)^{\frac{1}{2}}\\ \\ \frac{dD}{dx} &= \frac{x}{\sqrt{x^2 +4}}\\ \\ \\ \frac{dD}{dt} &= \left(\frac{dD}{dx} \right)\left(\frac{dx}{dt} \right )\\ \\ &= (\frac{x}{\sqrt{x^2 +4}})(500)\\ \\ \\ &= \left(\frac{500x}{\sqrt{x^2 +4}} \right)\\ \\ \text{When D} &= 5, x =\sqrt{21}\\ \\ \therefore \frac{dD}{dt} = \frac{(500)(\sqrt{21})}{\sqrt{(\sqrt{21})^2 +4}}\\ &= 458 kmhr" /></a>
<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{dx}{dt} &= 500 km h^{-1}\\ \\ D &= (x^2 @plus;4)^{\frac{1}{2}}\\ \\ \frac{dD}{dx} &= \frac{x}{\sqrt{x^2 @plus;4}}\\ \\ \\ \frac{dD}{dt} &= \left(\frac{dD}{dx} \right)\left(\frac{dx}{dt} \right )\\ \\ &= (\frac{x}{\sqrt{x^2 @plus;4}})(500)\\ \\ \\ &= \left(\frac{500x}{\sqrt{x^2 @plus;4}} \right)\\ \\ \text{When D} &= 5, x =\sqrt{21}\\ \\ \therefore \frac{dD}{dt} = \frac{(500)(\sqrt{21})}{\sqrt{(\sqrt{21})^2 @plus;4}}\\ &= 458 kmhr" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{dx}{dt} &= 500 km h^{-1}\\ \\ D &= (x^2 +4)^{\frac{1}{2}}\\ \\ \frac{dD}{dx} &= \frac{x}{\sqrt{x^2 +4}}\\ \\ \\ \frac{dD}{dt} &= \left(\frac{dD}{dx} \right)\left(\frac{dx}{dt} \right )\\ \\ &= (\frac{x}{\sqrt{x^2 +4}})(500)\\ \\ \\ &= \left(\frac{500x}{\sqrt{x^2 +4}} \right)\\ \\ \text{When D} &= 5, x =\sqrt{21}\\ \\ \therefore \frac{dD}{dt} = \frac{(500)(\sqrt{21})}{\sqrt{(\sqrt{21})^2 +4}}\\ &= 458 kmhr" title="\frac{dx}{dt} &= 500 km h^{-1}\\ \\ D &= (x^2 +4)^{\frac{1}{2}}\\ \\ \frac{dD}{dx} &= \frac{x}{\sqrt{x^2 +4}}\\ \\ \\ \frac{dD}{dt} &= \left(\frac{dD}{dx} \right)\left(\frac{dx}{dt} \right )\\ \\ &= (\frac{x}{\sqrt{x^2 +4}})(500)\\ \\ \\ &= \left(\frac{500x}{\sqrt{x^2 +4}} \right)\\ \\ \text{When D} &= 5, x =\sqrt{21}\\ \\ \therefore \frac{dD}{dt} = \frac{(500)(\sqrt{21})}{\sqrt{(\sqrt{21})^2 +4}}\\ &= 458 kmhr" /></a>