dr/dt dA/dt dy/dt dx/dt ??? Calculus...?

Question

So I'm still having a bit of trouble understanding the concept of this 'derivative with time' statement.

Take for instance this problem:
circle with radius r. If dr/dt = 4
find dA/dt when r = 3

So...you take pi(r)^2 and take its derivative.
Ok. So the hardest problem with me is understanding what I need to do first. What is it I AM LOOKING for...

or another instance is...
xy = 5
let dy/dt = 1
find dx/dt when x = 1

its all so cumbersome to me still.

Answer 1

You are given dr/dt = 4, so, taking the antiderivative,
r = 4t + ro (ro is just the radius at time = zero)
=>
A = pi * r^2 = pi(16t^2 +8t*ro+ro^2)
=>
d/dt A = pi * d/dt (16t^2 +8t*ro+ro^2)
=>
d/dt A = pi* (32t + 8*ro)
if by "r=3" you mean ro=3 then
=>
d/dt A = pi * (32t + 24)

...

For the second problem, dy/dt = 1 means y=t+yo
and xy=5 means y=5/x (except when x=0)
so, in general, 5/x = t + yo
=>
5/(t+yo) = x provided t+yo is not zero
so, dx/dt = 5(-1)[(t+yo)^-2] = -5[(t^2+2t*yo+yo^2)^-1]
when x=1 y=5, so yo=5-t.
thus, dx/dt = -5[(t^2+2t(5-t)+(25-10t+t^2))^-1]
dx/dt = -5 [ (t^2 + 10t - 2t^2 + 25 - 10t + t^2)^-1]
dx/dt = -5 [ 25 ^ -1]
dx/dt = -5/25
dx/dt = -1/5

Answer 2

y = cos(x) dy = -sin(x) * dx x = pi/6 dx = 2 dy = -sin(pi/6) * 2 = -(a million/2) * 2 = -a million dy = -a million V = (a million/3) * pi * r^2 * h dr = 2 h = 3r V = (a million/3) * pi * r^2 * (3r) = pi * r^3 dV = 3 * pi * r^2 * dr dV = 3 * pi * r^2 * 2 = 6 * pi * r^2 a) r = 6 dV = 6 * 36 * pi = 216pi b) r = 24 dV = 6 * 576 * pi = 3456pi we've a the excellent option triangle with a hypotenuse of 25 ft. x^2 + y^2 = 25^2 2x * dx + 2y * dy = 0 x * dx + y * dy = 0 dx = 2 2x + y * dy = 0 y * dy = -2x dy = -2x / y y = sqrt(25^2 - x^2) dy = -2x / sqrt(625 - x^2) dy = -2 * 7 / sqrt(625 - forty 9) = -14 / sqrt(576) = -14 / 24 = -7/12 ok, so while the backside of the ladder is 7 ft removed from the wall, and our substitute interior the area from the backside to the wall is +2 ft/s, then the substitute interior the top of the ladder is -7/12 ft/sec (minus 7 inches/2d). What we want is to discover the substitute interior the attitude that's made between the floor and the backside of the ladder. theta = arctan(y / x) = arctan(sqrt(625 - x^2) / x) The by-manufactured from arctan(m) = a million / (a million + m^2) m = sqrt(625 - x^2) / x) a million / (a million + (625 - x^2) / x) = a million / (x + 625 - x^2) / x) = x / (-x^2 + x + 625) x = 7 7 / (-forty 9 + 7 + 625) = 7 / (583) The angles is changing at 7/583 radians in keeping with 2d

Answer 3

First, a quick note: derivatives for this sort of problem may not necessarily be with respect to time - but usually are.

These problems are called related rates. The idea is that you have one quantity whose rate of increase or decrease is known, and you want to find the rate of change of some related quantity (hence the term related rates).

In your first example, we have a circle whose radius we are steadily increasing. The area of the circle is a function of the radius, so as we change the radius we will also change the area. A natural question to ask is how the change of radius will affect the area - or more precisely, if we change the radius at some particular rate, what will the rate of change of the area be?

All these problems work off the chain rule: dy/dx = dy/dz . dz/dx. In this case, we have dr/dt and we want to know dA/dt; we have an equation relating A to r. So the chain rule says
dA/dt = dA/dr . dr/dt
and from A = πr^2 we know dA/dr = 2πr. So we have
dA/dt = 2πr dr/dt
and now we substitute in the given values of r and dr/dt to get
dA/dt = 2π(3)(4) = 24π.

It quite often happens that we need to use implicit differentiation to get the missing piece (the equivalent of dA/dr in the last problem). In your second case this isn't really necessary, but we can do it that way to illustrate the point:

xy = 5
Differentiate with respect to x:
1.y + x.dy/dx = 0
=> dy/dx = -y/x
From the chain rule we have
dy/dt = dy/dx . dx/dt
and we are given dy/dt = 1, x = 1 so we get
1 = -y/x . dx/dt = -y dx/dt
=> dx/dt = -1/y
and at x = 1 we have y = 5 so we finish with dx/dt = -1/5.

With more experience you can save some steps by differentiating directly with respect to the dependent variable, e.g. in the last case
xy = 5, differentiate w.r.t. t:
dx/dt . y + dy/dt . x = 0
Substitute x = 1, dy/dt = 1 and y = 5/1 = 5 to get
dx/dt . 5 + 1 . 1 = 0
=> dx/dt = -1/5.

Or in the first problem:
A = πr^2
=> dA/dt = 2πr dr/dt
and substitute in r = 3, dr/dt = 4 to get
dA/dt = 2π(3)(4) = 24π.

To summarise:
You should have one rate that is known and one that you need to calculate. Identify these. Then find an equation linking the two quantities (area and radius, or x and y above). Use this to get the derivative of one quantity with respect to the other. Then use the chain rule to write one rate in terms of the other, using this derivative. Then substitute in known values and you can calculate the required rate.