A car is driving due north along a road while an airplane is flying to the east at an altitude of 4 miles from the intersection of north-east direction. How fast is the distance between car and airline changing when the car is 8 miles to the north and traveling at 80 mph, while the plane is flying at speed 220 mph and is at 12 miles from the intersection and gaining altitude at 5 mph?

Please solve in fractions, not decimals :)

To solve this problem, we can use the concept of related rates. We need to find the rate at which the distance between the car and the airplane is changing, which I will denote as "d'."

Let's denote the variables:

"x" will be the horizontal distance of the car
"y" is the vertical distance of the car from the intersection, which is measured northward
"d" is the distance the car is from the airplane

Using the Pythagorean theorem, we have...

[tex]d^2=x^2+y^2[/tex]

Differentiating both sides of the equation with respect to time, t, we get...

[tex]d^2=x^2+y^2\\\\\\\Longrightarrow 2d\cdot d'=2x \cdot x'+2y \cdot y'[/tex]

Solving for d'

[tex]2d\cdot d'=2x \cdot x'+2y \cdot y'\\\\\\\Longrightarrow \boxed{d'=\dfrac{xx'+yy'}{d} }[/tex]

Note that...

x'=0 mph because the car's horizontal distance is not changing
y'=80 mph which is the rate at which the car is moving north
x=12 mi
y=8 mi

Plugging these values in, we have...

[tex]d'=\dfrac{xx'+yy'}{d}\\\\\\\Longrightarrow d'=\dfrac{(12)(0)+(8)(80)}{d}; \ d=\sqrt{x^2+y^2}\\\\\\\Longrightarrow d'=\dfrac{(8)(80)}{\sqrt{x^2+y^2}}\\\\\\\Longrightarrow d'=\dfrac{640}{\sqrt{(12)^2+(8)^2}}\\\\\\\Longrightarrow d'=\dfrac{640}{4\sqrt{13}}\\\\\\\therefore \boxed{\boxed{d'=\dfrac{160\sqrt{13} }{13} \ mph}}[/tex]

Thus, the problem is solved.