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Calculus1 18 Online
OpenStudy (anonymous):

Joe is driving west at 60 km/h and Dave is driving south at 70 km/h. Both cars are approaching the intersection of the 2 roads. At what rate is the distance between the cars decreasing when Joe's car is 0.4 km and Dave's is 0.3 km from the intersection.

OpenStudy (anonymous):

oooh i bet they are \(.5\) miles apart then right?

OpenStudy (anonymous):

ok so i did this the way my teach taught me

OpenStudy (anonymous):

|dw:1431482336145:dw|

OpenStudy (anonymous):

what i did was 60 times 0.4 for joe and 70*0.3 for dave

OpenStudy (anonymous):

you want \(d'\) and you know \[x^2+y^2=d^2\] differentiate get \[2xx'+2yy'=2dd'\] or \[xx'+yy'=dd'\] you know all the numbers you need to solve this for \(d'\)

OpenStudy (anonymous):

for that 24(60)+70(21)=sqrt1017 dy/dt

OpenStudy (anonymous):

\[x'=60,y'=70,x=.4,y=.3,d=.5\]

OpenStudy (anonymous):

you lost me there with the square root

OpenStudy (anonymous):

you are given all the numbers you need to solve this equation \[xx'+yy'=dd'\] for \(d'\) which is what you are looking for

OpenStudy (anonymous):

well didnt i have to find d?

OpenStudy (anonymous):

well, you are not told \(d\) but you can find it with pythagoras

OpenStudy (anonymous):

yes i did that and got 31.89

OpenStudy (anonymous):

ooh no

OpenStudy (dan815):

there is a very simple geometric solution if youd like to see that

OpenStudy (anonymous):

\[\sqrt{.3^2+.4^2}=.5\]

OpenStudy (anonymous):

it is the world famous 3 - 4 - 5 mother of all right triangles

OpenStudy (anonymous):

|dw:1431482684493:dw|

OpenStudy (anonymous):

wait so their are two trianlges to this question ?

OpenStudy (anonymous):

no only one

OpenStudy (anonymous):

why cant i times 0.4 to 60 ?

OpenStudy (anonymous):

i was looking at my notes and the teacher did the similar way

OpenStudy (anonymous):

you are told how far away they are from the intersection right? you don't need to calculate it, you are told one is .3, the other is .4, so the distance between them is .5

OpenStudy (anonymous):

no doubt your teacher did a problem where you had to calculate how far they had gone, but in this case is it different you don't need to find how far they have gone, you are told how far they are from the intersection

OpenStudy (anonymous):

the problem i am refering too is somthing like " man goes south lady goes north , wahts the distance 2 mins later" so these tyoes are different ?

OpenStudy (anonymous):

yeah because in that one they are moving away from each other so you would need to find out how far they had gone

OpenStudy (anonymous):

so i did what u told me 21(70)+24(60)=0.5dx/dt? right?

OpenStudy (anonymous):

|dw:1431482974009:dw|

OpenStudy (anonymous):

no lets go slow

OpenStudy (anonymous):

you do not need to find x, you are told x, x is .3 you do not need to find y, you are told y, y is .4

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

\[70\times .3+60\times .4=.5d'\]

OpenStudy (anonymous):

i hope i got them in the right order, i forget check that it is right, the .3 goes with the 70 and the .4 with the 60 not the other way around

OpenStudy (anonymous):

yeah that is right

OpenStudy (anonymous):

dammnn thank you so much

OpenStudy (anonymous):

do u hate it when the notes dont match the HW and u get stuck ?

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