Math  /  Calculus

Question3. [0/1 Points] DETAILS MYNOTES
A plane flying horizontally at an altitude of 3 miles and a speed of 440mi/h440 \mathrm{mi} / \mathrm{h} passes directly over a radar station. Find the rate at which the distance from the plane to the station is increasing when it has a total distance of 4 miles away from the station. (Round your answer to the nearest whole number.)
292 xx mi/h
Enhanced Feedback Please try again. Keep in mind that distance =( altitude )2+( horizontal distance )2=\sqrt{(\text { altitude })^{2}+(\text { horizontal distance })^{2}} (or y2=x2+h2y^{2}=x^{2}+h^{2}.) Differentiate with respect to tt on both sides of the equation, using the Chain Rule, to solve for dydt\frac{d y}{d t}. The given speed of the plane is dxdt\frac{d x}{d t}. Need Help? Read It

Studdy Solution
Substitute the known values into the differentiated equation:
- y=4miles y = 4 \, \text{miles} - h=3miles h = 3 \, \text{miles} - dxdt=440mi/h \frac{dx}{dt} = 440 \, \text{mi/h}
First, find x x using the Pythagorean theorem:
42=x2+32 4^2 = x^2 + 3^2 16=x2+9 16 = x^2 + 9 x2=7 x^2 = 7 x=7 x = \sqrt{7}
Now substitute into the differentiated equation:
2(4)dydt=2(7)(440) 2(4) \frac{dy}{dt} = 2(\sqrt{7})(440)
Solve for dydt \frac{dy}{dt} :
8dydt=8807 8 \frac{dy}{dt} = 880 \sqrt{7} dydt=88078 \frac{dy}{dt} = \frac{880 \sqrt{7}}{8} dydt=1107 \frac{dy}{dt} = 110 \sqrt{7}
Calculate 1107 110 \sqrt{7} to the nearest whole number:
1107290.5 110 \sqrt{7} \approx 290.5
So, rounding to the nearest whole number, the rate at which the distance is increasing is:
291mi/h \boxed{291 \, \text{mi/h}}

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