anonymous
  • anonymous
A car is traveling east at 100 mph and a truck is traveling north at 80 mph. when the car is 3 miles east of an intersection and the truck is 4 miles north of the same intersection, how fast is the distance between them changing?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
@DanJS
DanJS
  • DanJS
this is one of those related rates probs
DanJS
  • DanJS
|dw:1449351226767:dw|

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More answers

DanJS
  • DanJS
It is a right triangle, and both sides are growing at different rates
DanJS
  • DanJS
write down everything that is given first... let the car distance be x, and the truck be y.. |dw:1449351636718:dw|
DanJS
  • DanJS
y = 4 mi x = 3 mi dy/dt = 80 mph dy /dt = 100 mph FIND --- dh / dt, when x and y are those values
DanJS
  • DanJS
the relationship for the three sides is the pythagorean theorem x^2 + y^2 = h^2
DanJS
  • DanJS
take the derivative of that w.r.t time (d/dt)
DanJS
  • DanJS
have to use the chain rule, implicit differentiaon 2*x*(dx/dt) + 2*y*(dy/dt) = 2*z*(dz/dt)
DanJS
  • DanJS
you can calculate z from the pythagorean theorem , for when x=3 and y=4, tha tmoment in time
DanJS
  • DanJS
so the hypotenuse is 5 put all the things in, the only thing unknown is dz/dt, how fast the hypotenuse is changing at that instance in time
DanJS
  • DanJS
2*x*(dx/dt) + 2*y*(dy/dt) = 2*z*(dz/dt) 2**3*100 + 2*4*80 = 2*5*[dz/dt]
DanJS
  • DanJS
Overall- you have a right triangle with 2 of the sides growing longer at different rates they want you to figure the info for the third side at a certain instance in time
DanJS
  • DanJS
take a derivative w.r.t time t that relates all the rates for each side, the only thing unknown is rate dz/dt
DanJS
  • DanJS
Givens - right-triangle (north and east) y = 4 mi x = 3 mi dy/dt = 80 mph dx /dt = 100 mph FIND --- dh / dt
DanJS
  • DanJS
yeah the chain rule thing, implicit differentiation is like this here x,y,and z are all functions of time t, distance traveled = rate*time
DanJS
  • DanJS
if you want to take a derivative of x w.r.t. time t, \[\large \frac{ d }{ dt }* x\] apply the chain rule \[\frac{ d }{ dt }[x] = \frac{ d }{ dx }*\frac{ dx }{ dt }[x]\]
DanJS
  • DanJS
i have to run for a bit, i will do the next one in a few
anonymous
  • anonymous
okay thanks.
DanJS
  • DanJS
to start if you want maximum area of a cylinder , when it has a certain surface area write the formulas for the area and surface area you want to maximize the area, have to change it to one variable by solving the surface area for radius or height, and subbing that into the area A(h) or A(r) take A '(h) or A ' (r) the max is at a critical point when A '= 0
DanJS
  • DanJS
remember doing maximum and minimum values of functions, the graph changes increasing/decreasing there... first derivative is zero The area function will have max when derivative of area function is zero
DanJS
  • DanJS
solve S.A for h sub h = into the area you have A(r) area as a function of radius , take derivative and find when 0

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