Question 1
Water is being drained from a pond such that the volume V (in m^3) of water in the pond after t hours is given by V = 5000(60 - t)^2. Find the rate at which the pond is being drained after 4 h.
Question 2The velocity of an object moving with constant acceleration can be found from the equation v = (v[0] ^2 + 2as)^2, where v[0] is the initial velocity, a is the acceleration, and s is the distance traveled. Find dv/ds.
Example 3The electric field E at a distance r from a point charge is E = k/r^2, where k is a constant. Find an expression for the instantaneous rate of change of the electric field with respect to r.
Example 4The distance s (in m) traveled by a subway train after the brakes are applied is given by s = 20t - 2t^2. How far does it travel, after the brakes are applied, in coming to a stop?
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Question #1:
ReplyDeleteV = 5000 (60 - t)^2
let,
u = 60 - t
therefore,
du/dt = -1
now,
dv/du = 10,000 (u)
= 10,000 (60 - t)
using product rule,
i.e. dv/dt = dv/du * du/dt
we get;
dv/dt = 10,000 (60 - t) . -1
hence,
the rate of change of V when t = 4 is:
dv/dt = 10,000 (60 - 4) . -1
= -560,000 m/s
QUESTION 1
ReplyDeleteV= 5 000 (60-t)^2
dV/ dt = 1 0000(60-t)
substitute t=4
dV/ dt = 10 000(60-4)
= 10 000 (56)
= 56 0000 m^3/s
wel thats what i thought it was.....can any1 else explain?
This comment has been removed by the author.
ReplyDeleteQUESTION 1
ReplyDeleteIf V= 5000 (60-t)^2: where V(volume of water in pond: m^3); t(time: hrs).
The question asks for "the rate" at which the pond at which the pond is being drained at 4 hours..........?
As soon as a math problem demands for "RATE"
This implies the derivative (differential) of the problem is wanted...........
So w.r.t. the variables given in question 1:
dV/dt= [(5000)(2)]*(60-t)^(2-1)=dy/dx= na*x^(n-1)
(diff. of y= ax^n)
dV/dt= 10 000*(60-t)
So THE RATE (dV/dt) at which the pond is being drained is: 10 000*(60-t).
Recalled that "t" is the time taken (hrs).......
So THE RATE at which this occurs for 4 HOURS is:
We simply SUBSTITUTE 4 in the equation we now found.......
So, dV/dt= 10 000*[(60-4)]
= 10 000*56
=560 000 m/s
Ans= 560 000 m/s
EXAMPLE 3:
ReplyDeleteIf E= k/r^2: where k is a CONSTANT
Find an expression for the "INSTANTANEOUS RATE OF CHANGE OF THE ELECTRIC FIELD" w.r.t. r.......
So the problem is demanding the differential of E w.r.t. r
i.e. dE/dr.
But before we jump into that, lets rearrange our ORIGINAL eq'n in the form: y= ax^n........so it will be easier to differentiate.
If E= k/r^2
E= kr^-2...(1)denom. is multiplied by num. & (2)exponent (power) becomes negative!!!!
So E= kr^-2 is in the form, y= ax^n: where, [E=y k=a, r=x and, -2=n]
So now, dE/dr= [(-2)k]*r^[-2-1]....as you would see from my last comment on how to diff. this term
dE/dr= -2kr^-3
Answer= -2kr^-3
que 2
ReplyDeletethe keys ponits to note in this ques while reading is velocity and accelaration. becasuse they both are rate of speed over time and velocity over time. hmm this ques is kinda tricky but my guess is you might have to differentiate the eqn v = (v[0] ^2 + 2as)^2
the way in which nerd worked out question on was understandable because it was explaned properly because i cant really understand what phoenix doing because it was not explaned
ReplyDeleteGlad u understood clearly renjisan!!!!!!!
ReplyDeleteI totally agree with nerd he couldn't have put it across better.
ReplyDeletenerd is a bright one boy yea nerd i understand wat u do
ReplyDeleteWell renjisan, ridiclyric and Hu$tler, let me read you explain what nerd said in YOUR words
ReplyDeletefrom question 1 most of the question is just background. what you should be concerned with is the formula they gave you and the volume with respect to time dV/dt. when your done differentiating, you jus substitute the value of t given . I find the differentiation is very easy once you're given the formula.
ReplyDeletei agree with phoenix with substituting the expression in the bracket with a variable and differentiate the variable with respect to the unknown number of the term. you then differentiate the original equation with respect to the substituted variable. i'm just trying to put into words what phoenix did as the solution
ReplyDeletei think this is how no 1 is done:
ReplyDeleteV = 5000(60-t)^2
= 5000(3600 - 120t + t^2)
= 18000000 - 600000t + 5000t^2
dv/dt = 10000t - 600000
at t=4
dv/dt = 10000(4) - 600000
= -560000 m^3/hr
answer for miss W.R.T ques 1
ReplyDeleteok well first he starts out by putting the eq'n that was given. and he goes on to say that as soon as u see rate you know that there is some differentiation involed and then he metioned that u differentiate w.r.t the variables which is V volume and T time.
after he differentiated and got an expression, he goes on to say time(t) is in hours and the ques gave t=4 so he simply substituded t into the expression and got that answer
How to work out this question
ReplyDeletes = 2t^3 -4t^2 ; t = 4
Ok darky, am willin 2 help:
ReplyDeleteSo u got: s= 2t^3 - 4t^2; t=4:
STEP 1: We need 2 find the differential of s w.r.t. t, i.e. ds/dt.....cool?
So diff. of 's' w.r.t. t is ds/dt:
So, ds/dt= [(3)(2)*t^(3-1)] - [(4)(2)*t(2-1)]
ds/dt= 6t^2 - 8t
STEP 2:
SUBSTITUTE t=4 into ds/dt, i.e the eq'n we now derived.
When, t=4:
ds/dt= (6)[(4)^2] - 8(4)
ds/dt= (6)(16) - 32
ds/dt= 96-32
ds/dt= 64
Ans.= 64.
VERY IMPORTANT: To differentiate an expression in the form:
y= ax^n.
We get: dy/dx= n*a*x^(n-1).........n frm the power is multiplied by the coefficient (number in front of the variable, x in this case) of the variable; x and, THE POWER (n is this case) IS REDUCED BY 1. Hence the diff. of an eq'n in the form, y= ax^n= dy/dx= n*a*x^(n-1)
that's a really good explanation nerd
ReplyDeleteExample 4.
ReplyDeleteThe dist. travelled is given by :
s = 20t - 2t^2
the next thing to do is find out what the ques is askin for.
I think it is askin for the rate of change i distance w.r.t time
ie. ds/dt
therefore:
ds/dt = [ 1*20t^(1-1)] - [2*2t^(2-1)]
ds/dt = 20 - 4t
note: the ques didnt ask for an instantaneous answer...
if im wrong cud someone pls correct me...
ur correct with the answer Dark Angel. after reading the question i interpret it to be asking for the change in distance w.r.t time ds/dt. so you are correct to jus differentiate the equation
ReplyDeleteQ2
ReplyDeletev = (v[0] ^2 + 2as)^2
where v(o)is initial velocity
a is the acceleration
s is distance traveled
rtf
dv/ds
so therefore:
ok 1st let (v[0] ^2 + 2as) be u
so wen
v= u^2
dv/du is = 2( u^2-1 )
dv/du = 2u
and
u = { v(0)^2 + 2as }
du/ds = 0 + 2as^1-1
du/ds = 2a
therefore:
dv/ds = du/ds * dv/du
dv/ds = 2a * 2u
subst back 4 u :
dv/ds = 2a { v(0)^2 + 2as }
dv/ds = 2v(0)^2 + 4 a^2 s
question#1
ReplyDelete(60-t)^2 let y = 60-t
du/dt=1
let v = u^2
d^2/du= 2u
du/dt x dv/du
=dvdt
=2u x 1
=24 x 1
=24 x 5000
=10,000u
10,000(60-t)
when t=4
10,000(60-4)=560,000^3m per. hr.
Given
ReplyDeleteE = k/r^2, where k is a constant
let
E = k ( r^-2 )
cause u have 2 put everything on 1 line
dE/dr= -2(k) r^(-2-1)
dE/dr= -2 k r^(-3)
dE/dr= -2k/r^3
hey does anyone know how to find the distance in number 4, because a particular time isnt given????
ReplyDeleteConcept of differentiation
ReplyDeleteIn mathematics, a fundamental concept of differential calculus representing the instantaneous rate of change of a function. The first derivative of a function is a function whose values can be interpreted as slopes of tangent lines to the graph of the original function at a given point. The derivative of a derivative (known as the second derivative) describes the rate of change of the rate of change, and can be thought of physically as acceleration. The process of finding a derivative is called differentiation.
What is differentiation ?
ReplyDeleteDifferentiation is a method to compute the rate at which a dependent output y, changes with respect to the change in the independent input x. This rate of change is called the derivative of y with respect to x.
ReplyDeleteThe questions above are examples of differentiation since they asked to find a measure of how a function changes as its input changes.
ReplyDeleteFor example, the derivative of the position (or distance) of a vehicle with respect to time is the instantaneous velocity (respectively, instantaneous speed) at which the vehicle is traveling. Conversely, the integral of the velocity over time is the vehicle's position.
Today's question deals with differentiation and spheres. The question uses some very nice properties of differentiation and the dy/dx notation.
ReplyDeleteA sphere has radius of length r cm, surface area of S cm2, and a volume of V cm3 on a given instant, t.
1) Prove: (dV/dt) * (dr/dt) = (dS/dt)2 / 16π
2) Find the surface area when dV/dt = π cm3/sec and dS/dt = 2π cm2/sec
Not that bad, is it?
First, there are two important formulas we are going to use in order to solve this problem:
V = 4πr3 / 3
S = 4πr2
Now differentiate each function according to t:
dV/dt = 4πr2 dr/dt
dS/dt = 8πr dr/dt
To get dV/dt * dr/dt, multiply the first equation by dr/dt:
dV/dt * dr/dt = 4πr2 (dr/dt)2
Now square the second equation:
(dS/dt)2 = 64π2r2 (dr/dt)2
Divide the second equation by 16π:
64π2r2 / 16π = 4πr2 (dr/dt)2
Now this is exactly the same as dV/dt * dr/dt.
Q.E.D
The second question:
dS/dt = dS/dV * dV/dt
dS/dV = dV/dt * dS/dt
dS/dV = (8πr dr/dt) / (4πr2 dr/dt)
dS/dV = 2/r
dS/dt = 2/r * dV/dt
dS/dt = 2/r * π
dS/dt = 2π
2π = 2π/r
1 = 1/r
r = 1
Plug r = 1 into the surface area function:
S = 4π * 12 = 4π cm2
Problem solved.
Question 1
ReplyDeleteWater is being drained from a pond such that the volume V (in m^3) of water in the pond after t hours is given by V = 5000(60 - t)^2. Find the rate at which the pond is being drained after 4 h.
V = 5000 (60 - t) ^2
In order to find the rate at which the pond is being drained after 4 hours:
Firstly:
we need to find V with respect to time which is
dV/dt.
And to find this we use the differentiation rule "y = ax^n so dy/dx = anx^n-1"
But since the formula given does not look like this but has to be differeniated using this rule we make a substitution.
so....
let s = 60 - t.
This also has to be differentaited using the same rule
ds/dt = -1
Inwhich the equation now looks like
V = 5000s^2
dV/ds = 1000s
now rewrite to find dV/dt
dV/dt = ds/dt * dV/ds
= 10000s * -1
but s = 60 - t
so
dV/dt = 10000(60 - t) * -1
Secondly to find the rate at 4 hrs, just simply substitue t=4 in the equation for dV/dt
Question 2
ReplyDeleteThe velocity of an object moving with constant acceleration can be found from the equation v = (v[0] ^2 + 2as)^2, where v[0] is the initial velocity, a is the acceleration, and s is the distance traveled. Find dv/ds.
v = (v[0] ^2 + 2as)^2
we all know that 1 is understood in front of
(v[0] ^2 + 2as)^2
so the rule for this differentiation is
"y = ax^n so dy/dx = anx^n-1"
However we make a substitution
let p = v[0]^2 + 2as
dp/ds = 2v[0] + 2
Rwrite
v = p^2
dv/dp = 2p
Therefore
dv/ds = dp/ds * dv/dp
= 2v[0] + 2 * 2p
remember that p = v[0]^2 + 2as
dv/ds = v[0]^2 + 2as * 2v[0] + 2
Question 3
ReplyDeleteThe electric field E at a distance r from a point charge is E = k/r^2, where k is a constant. Find an expression for the instantaneous rate of change of the electric field with respect to r.
E = k/r^2
Make the equation simplier
E = Kr^-2
dE/dk = -2Kr
Question 4
ReplyDeleteThe distance s (in m) traveled by a subway train after the brakes are applied is given by s = 20t - 2t^2. How far does it travel, after the brakes are applied, in coming to a stop?
s = 20t - 2t^2
ds/dt = 20 - 4t
How far it travels really means that they want the maximum point which is when ds/dt = 0
0 = 20 - 4t
4t = 20
t = 20/4
t = 5