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Mathematics 18 Online
OpenStudy (anonymous):

some calculus anyone? see pic:

OpenStudy (anonymous):

im dying!

OpenStudy (watchmath):

interesting problem. I'll come back later if nobody help you :).

OpenStudy (anonymous):

maybe i should write my progress too \[\tan \Theta =((4m \sin \alpha) \div (m + 4 m \cos \alpha)) +1\] for q a

OpenStudy (amistre64):

the velocity vector of the truck is 4 times the magnitude of the car then right?

OpenStudy (amistre64):

add the vectors to get the new vector direction

OpenStudy (amistre64):

< 0, y> <4x,4y> -------- <4x, 5y> right?

OpenStudy (amistre64):

that should be <x,0>

OpenStudy (amistre64):

new vector is <5x,4y> then

OpenStudy (anonymous):

yeah, now im with you...

OpenStudy (amistre64):

whenim wring let me know lol

OpenStudy (anonymous):

i had to look at the question again :)

OpenStudy (amistre64):

the new angle they go off in is y over x

OpenStudy (amistre64):

4y ---- = tan(t) , but how to but that in sin cos? 5x

OpenStudy (amistre64):

y = sin ; and x = cos right?

OpenStudy (anonymous):

yeah so \[4\sin (\alpha)/4\cos(\alpha)= \tan(\theta)\]

OpenStudy (amistre64):

truck vector = <4cos(a),4sin(a)> car vecotr = <1,0>

OpenStudy (amistre64):

add them to get: < 1 , 0 > <4cos(a),4sin(a)> --------------- <4cos(a) +1, 4sin(a)> = new direction

OpenStudy (anonymous):

oh, my numer and denom should be reversed

OpenStudy (amistre64):

which is: tan(t) = 4sin(a)/(4cos(a)+1)

OpenStudy (amistre64):

b wants you to tell them the velocity, the magnitude, that this new vector is i think

OpenStudy (amistre64):

try using distance formula with new vector to see

OpenStudy (anonymous):

but tan =cos/sin not tan = sin/cos

OpenStudy (amistre64):

scalar the inital velocity of each to u perhaps?

OpenStudy (amistre64):

tan = sin/cos ; always has and always will

OpenStudy (amistre64):

tan = slope = y/x = sin/cos

OpenStudy (anonymous):

dam i looked at cot

OpenStudy (anonymous):

bjargh all i can get is \[d=\sqrt{16\cos^2(\alpha) + 16 \sin^2(\alpha)}\]

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