1. ## Related Rates Questions..

A car traveling at a rate of 30 ft/sec is approaching an intersection. When the car is 120 ft from the intersection, a truck traveling at a rate of 40 ft/sec crosses the intersection. If the roads are at right angles to each other, how fast are the car and the truck separating 2 seconds after the truck crosses the intersection?

.If a ladder of length 30 ft that is leaning against a wall has its upper end sliding down the wall at a rate of 1/2 ft/sec, what is the rate of change of the acute angle made by the ladder with the ground when the upper end is 18 ft above the ground?

2. Originally Posted by Ajohnson121
A car traveling at a rate of 30 ft/sec is approaching an intersection. When the car is 120 ft from the intersection, a truck traveling at a rate of 40 ft/sec crosses the intersection. If the roads are at right angles to each other, how fast are the car and the truck separating 2 seconds after the truck crosses the intersection?
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I'm going to think of the car as traveling from east (-) to west (+) and the truck from south (-) to north (+). First think how these movements are related. Since the roads are at right angles, we can use a right triangle and the pythagorean theorem. Let's call the car's road a, and the truck's road b. The distance between them we'll call c. We need to know how fast they are separating, so we need to know how c is changing over time or $\displaystyle \frac{dc}{dt}$

$\displaystyle a^2+b^2=c^2$

To get $\displaystyle \frac{dc}{dt}$ we need to take the derivative with respect to t.

$\displaystyle 2a*\frac{da}{dt}+2b*\frac{db}{dt}=2c*\frac{dc}{dt}$

Now plug in the information that you know. Note that the car has moved from 120 ft away to 60 ft away in the 2 seconds, etc.

$\displaystyle 2*(-60)*30+2*80*40=2*\sqrt{40^2+(-60)^2}*\frac{dc}{dt}$

Solve for $\displaystyle \frac{dc}{dt}$