Calculus - Related Rates: From a certain spot X, a car travels due north at a speed of 70 miles per hour. Ten miles east of this spot X, another car travels due south at a speed of 80 miles per hour. How fast is the distance between the two cars increasing the moment they are 40 miles apart?
differentiate \[x^2+y^2=z^2\] with respect to time t
2x dx/dt + 2y dy/dt =2z dz/dt
good...did you draw a picture...what is x ,y, and z
|dw:1318795515867:dw|
we can then make a large triangle...
|dw:1318795596555:dw|
I guess I shouldn't have used X since they are referring to it as a point
i'm with you so far
y=10,z=40...find x
x= \[\sqrt{1500}\]
sorry dude, this script is killing my ie8. had to bail and switch to mozilla.
do you know what x' and y' are?
yes, x' is 70mph and y' is 80mph
no
y is a fixed 10 miles...so y'=0
ah hah
and x' is not 70
north and south are both vertical displacements so x'=70+80=150
ok, gotcha
plug in all your numbers into your derivative and solve for z'
145.24mph?
yes
awesome, thank you for your help. I was doing some crazy stuff I didn't need to do.
np
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