Question

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?

Answer 1

@DanJS

Answer 2

this is one of those related rates probs

Answer 3

It is a right triangle, and both sides are growing at different rates

Answer 4

write down everything that is given first...

let the car distance be x, and the truck be y..

Answer 5

y = 4 mi
x = 3 mi

dy/dt = 80 mph
dy /dt = 100 mph


FIND --- dh / dt, when x and y are those values

Answer 6

the relationship for the three sides is the pythagorean theorem


x^2 + y^2 = h^2

Answer 7

take the derivative of that w.r.t time (d/dt)

Answer 8

have to use the chain rule, implicit differentiaon

2*x*(dx/dt) + 2*y*(dy/dt) = 2*z*(dz/dt)

Answer 9

you can calculate z from the pythagorean theorem , for when x=3 and y=4, tha tmoment in time

Answer 10

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

Answer 11

2*x*(dx/dt) + 2*y*(dy/dt) = 2*z*(dz/dt)

2**3*100 + 2*4*80 = 2*5*[dz/dt]

Answer 12

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

Answer 13

take a derivative w.r.t time t

that relates all the rates for each side, the only thing unknown is rate dz/dt

Answer 14

Givens -
right-triangle (north and east)
y = 4 mi
x = 3 mi
dy/dt = 80 mph
dx /dt = 100 mph


FIND --- dh / dt

Answer 15

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

Answer 16

if you want to take a derivative of x w.r.t. time t,

\[\large \frac{ d }{ dt }* x\]

apply the chain rule

\[\frac{ d }{ dt }[x] = \frac{ d }{ dx }*\frac{ dx }{ dt }[x]\]

Answer 17

i have to run for a bit, i will do the next one in a few

Answer 18

okay thanks.

Answer 19

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

Answer 20

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

Answer 21

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