application of calculus to the physical help (1 Viewer)

[wrong_turn]

i find that i dont really understand this topic very well. or not at all get it as well. can anyone give me tips on how to understand this topic in regards to the sub-topics within this topic?

i also have trouble with this question.

the rate of chage in the radius of a sphere is 0.3mms^-1. given that the radius is 88mm at a certain time, find the rate of chage at that time in the
a) surface area
b) volume

 

[SaHbEeWaH]

sit and spend time learning the concepts behind it that's all

you're given dr/dt,
and they're asking for the rate of change of surface area when r = 88,
i.e. dA/dt when r = 88
now just recall the surface area of a sphere, 4 pi r² or something like that...
now that you have A in terms of r
you can differentiate and get dA/dr and use the knowledge that dA/dr * dr/dt = dA/dt...

works in a similar fashion with volume

 

[FinalFantasy]

i find that i dont really understand this topic very well. or not at all get it as well. can anyone give me tips on how to understand this topic in regards to the sub-topics within this topic?

i also have trouble with this question.

the rate of chage in the radius of a sphere is 0.3mms^-1. given that the radius is 88mm at a certain time, find the rate of chage at that time in the
a) surface area
b) volume

a)dr\dt=0.3mms^-1
surface area of the sphere is S=4pi*r²
dS\dr=8pi*r
rate of change of surface area is:
dS\dt=dS\dr*dr\dt
dS\dt=8pi*r dr\dt
at the time the radius is 88mm:
dS\dt=8pi(88)*0.3 mm²\s
b) volume of the sphere is V=(4\3)pi*r³
dV\dr=4pi r²
dV\dt=dV\dr* dr\dt
dV\dt=4pi r² dr\dt
at that time, radius is 88mm:
dV\dt=4pi(88)²* 0.3 mm³\s

if u don't understand the topic just read the examples provided in text books and ask what u don't understand

 

[Jago]

what's the difference between 2u and 3u rates of change?

 

[acmilan]

3 unit rates of chain would undoubtedly involve the chain rule in one way or another.

eg they may say find the rate at which the area of a circle is increasing given that the radius is increasing at a rate k m/s. So you need to find dA/dt, but A is not in terms of t, hence you must use dA/dt = dA/dr * dr/dt, where dr/dt would be given to you. Meanwhile in 2unit, they may say something like find the rate at which the area of a circle is increasing with respect to its radius. In that case, all you have to do is find dA/dr by differentiating the formula for area, A = pi*r², giving dA/dr = 2pi*r

 

[pikto]

velocity = a (variable)u (units)^-1 = dy/dx of original eqn.
acceleration = au^-2 = d^2y/dx^2 of original eqn.
'initially' t (time) = 0
'at the origin' s (displacement)(original eqn) = 0
i know you have seen this all before, but this is pretty much (with the exception of exponentials) it. keep in mind that if finding the eqn of displacement, it is probably an indefinite integral (+c), so note all clues as to values in the question. NB: as in any calculus application all of the actions are reversible via integration.
seriously, this stuff is so easy, once you get it you will cry.
(i tried to think of it as much the same as maxima/minima)

 

[Dr_Gorgeous]

*whoosh* that went straight over my head! nah, we had a test on it yesterday...did pretty good (meaning i'll probably get 40%)