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# Related rates: Approaching cars

AP.CALC:
CHA‑3 (EU)
,
CHA‑3.E (LO)
,
CHA‑3.E.1 (EK)

## Video transcript

so this car right over here is approaching an intersection at 60 miles per hour and right now right at this moment it is 0.8 miles from the intersection now we have this truck over here it's approaching the same intersection on a street that is perpendicular to the street that the car is on and right now it is 0.6 miles so it is 0.6 miles from the intersection and is approaching the intersection at 30 miles per hour now my question to you is what is the rate at which the distance between the car and the truck is changing well to think about that let's first just think about what we're asking so we're asking about the distance between the car and the truck so right at this moment when the car is 0.8 miles from the intersection the truck is 0.6 miles from the intersection the truck is traveling at 30 miles per hour towards the intersection the car is traveling 60 miles per hour towards the intersection right at this moment what is the rate at which this distance right over here is changing and just so that we have some variables in place let's call this distance s so what we really are trying to figure out is right at this moment what is DS DT going to be equal to well let's think about what we know that we could use to somehow come to terms or figure out what DS DT is well we know we know the distance of the car and the intersection and let's just call that distance let's call that I don't know let's call that distance Y so Y is equal to 0.8 miles we also know that D so let me write this we know that Y is 0.8 miles right now 0.8 miles we also know that dy DT the rate at which Y is changing with respect to time is what well Y is decreasing by 60 miles per hour so let me write it as negative 60 miles miles per hour now similar similarly let's say that this distance right over here is X is 0.6 miles right at this moment so we know that X is equal to 0.6 miles what is the rate at which X is changing with respect to time well we know it's 30 miles per hours how fast we're approaching the intersection but X is decreasing by 30 miles every hour so we should say it's negative 30 miles miles per hour so we know what Y is we know what X is we know how fast Y changing how fast X is changing with respect to time so we could try to do here is come up with a relationship between X Y and s and then may differentiate that relationship with respect to time and it sounds seems like we have pretty much everything we need to solve for this so it's a relationship between X Y and s well we know that this is a right triangle the streets are perpendicular to each other so we can use the Pythagorean theorem we know that x squared plus y squared is going to be equal to s squared and then we can take the derivative of both sides of this with respect to time to get a relationship between all the things that we care about so what's the derivative of x squared with respect to time well so you're going to be the derivative of x squared with respect to X which is just 2 x times the derivative of X with respect to time x dx/dt once again just the chain rule derivative of something squared with respect to the something times the derivative of the something with respect to time and we use similar logic right over here when we want to take the derivative of Y squared with respect to time derivative of Y squared with respect to Y times the derivative of Y with respect to time now on the right hand side of this equation we once again take the derivative with respect to time so it's the derivative of s squared with respect to s which is just 2 s times the derivative of s with respect to time once again this is all just an application of the chain rule so now it looks like we know what X is we know what DX DT is we know what Y is we know dy DT is all we need to figure out is what s and then what DS DT is the rate at which this distance is changing with respect to time well what s right now well we can actually use the Pythagorean theorem at this exact moment we know we know that x squared so X is zero point six we know zero point six squared plus y squared zero point eight squared is equal to s squared well this is 0.36 plus zero point six four is equal to s squared this is one is equal to s squared and we only care about positive distances so we have s is equal to one right now so we also know what s is so let's substitute all of these numbers in and then try to solve for what we came here to do solve for d s DT so the rate at which so 2 times X let me do that in yellow two times X X is 0.6 is going to be 1.2 times dx/dt so that's negative 30 miles per hour so times negative 30 miles per hour plus 2 times y is 1.6 1.6 times dy DT is negative 60 miles per hour negative 60 miles per hour and I'm not writing the unit's here but if you were to write the unit's you will see that all of our distances are in miles and all of our time is within hours so we're going to get an answer when we solve for d s DT that's miles per hour but I encourage if you if you want to to actually write out the unit's and see how they work out and so this is going to be equal to 2 times s well s is one mile so it's just going to be two times d s DT which is what we're trying to solve for so what do we get here on the left hand side so 1.2 times negative 30 that's negative 36 right 1/5 of 30 is 6 yep that's right and then 1.6 times negative 60 that's going to be negative 96 is equal to 2 times d s DT is equal to 2 times the rate at which our distance is changing with respect to time on the left hand side right over here this is negative 132 negative 132 is equal to 2 times d s DT divide both sides by 2 we get negative 66 and now we can put our units if we want miles per hour is the rate at which our distance is changing with time so DST T is negative 66 miles per hour does it make sense that we got a negative number here well sure this distance is decreasing and right at this moment as they approach the intersection
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