Differentiation - Practical Applications

[TwinkleTits]

Heres some practical questions. I'm not sure if I know how to do them though

1) The rate at which water is flowing in a valve is R cm^3 s
The flow of water varies with time, t seconds, as given by equation:
R = 8t - t^2

FIND:
i) dR/dt, the rate of change of R with respect to t
For this one is the answer just the differential eg. 8 - 2t?

ii) the value of dR/dt after 6 seconds
I'm guessing I plug 6 into the differentiated equation

the time when the rate of flow is a maximum and calculate the maximum
No idea about this one Im reading about maxima and minima at the moment

[Mathematics]

Sorry for the post. Somehow I managed to convince myself that the stationary point was at t < 0.

But Mantikore is absolutley right.

R(t)=8t-t^2

\partial_t (8t-t^2)=8-2t=0

at the stationary point.

So the maximum is at t=4

The maximum flow rate is then simply R(4).

Mantikore is also right about using the second derivative to check if a stationary point is a maxima or a minima, but in the case of quadratics you can just look at the sign of the quadratic term.

[LiquidIce]

What "level" is this? What grade and such? I'm just interested, wanted to compare my maths hl course to someone else ;)

[Mantikore]

thats really strange: maxima and minima is simple high school math, but you also seem to be doing complex numbers and matrix operations; a university level subject. i dunno, maybe the education system is different. anyway

notice that when you graph a function, when the curve reaches a certain peak or trough, the gradient at that point is zero. These "stationary points" are the maxima and minima respectively. A function may have multiple local maxima and minima and in some cases (like sin and cos), have infinite.

Anyway, youre derivative is 8-2t. the stationary point where the derivative is zero

8-2t = 0..................simple arithmetic here
t = 4
now usually, you have to be careful here. if there are multiple stationary points, they can either be a minimum, maximum, or inflex. The inflex is that bit around the origin in y=x^3. its neither a max or min, but its stationary point is still 0.

you can test the nature of this in a few ways:
1)draw the original function.
R=8t-t^2 is obviously a parabola. factorise it and you get R=t(8-t). the x-intercepts are obviously 0 and 8 and the parabola is upside down (upside down suggests that it is a maximum anyway). you can see that at t=4, R > 0, so its a maximum.similarly, if its a trough (valley like thing), its a minimum, and if its flat, then its an inflexion.

2)second derivative.
this is the more formal way. differentiate the derivative a second time. the derivative was linear before, but now its a constant.
d2y/dt^2 = -2.
luckily for you, its a constant, it will always be less than zero. this tells you its negative at t=4 (well, since theres no t's in there, its always negative :p), then its gotta be a maximum.

[Schwexx]

Yeah I remember in high school math doing similar max/min questions. Everytime we just went around solving them by taking the derivative and setting it to 0 to account for the peak.

then in first year calculus at university prof taught us it was entirely wrong and how to do it properly because it doesn't account for vertical tangents, corner points etc.

gotta love the education system :)

[redrussia]

You don't exactly need to test the case of possible maxima/minima for critical points (when the derivative = 0) because R(t) is a quadratic equation that is negative for the second degree polynomial, meaning that it only has one extreme, which is its maximum.

[KaBoom]

thats really strange: maxima and minima is simple high school math, but you also seem to be doing complex numbers and matrix operations; a university level subject. i dunno, maybe the education system is different. anyway

ive done max/min in standard AS Maths, complex numbers and matrix calc are in AS Further Maths. AS qualifications are generally for 16-17 year olds in the UK.