Question

hey! quick question.... how does implicit differentiation help simply work in a related rates problem?

Answer 1

dy/dt = dy/dx . dx/dt

Answer 2

So for example you are given y in terms of x and you are asked to find rate of change of y at a particular time, given rate of change of x that particular time, you can use the above relationship (implicit differentiation) to form an equation dy/dt = f(x,y,t) and just plug in the value for dx/dt, x, y and t to find the rate of change of y.

Answer 3

suppose boat A is moving horizontally at 6m/s east and boat B was moving towards the south at some speed. find that speed if after 10 seconds boat A is 100 meters from boat B;
you might come up with the following equation;
x^2+y^2=s
x^2+y^2=100

where x is the horizontal distance and y is the verticle distance
implicitly differentiating gives

2xx'+2yy' = 0
2x+2y(dy/dx) = 0 {x' = dx/dx}
-x/y = dy/dx

after 10 seconds x= 6*10 = 60meters
and y = sqrt(10000-3600) = 80meters

-60/80 = dy/dx

or you could use a ladder

Answer 4

thanks a lot!