A psycho is stabbing a person with a knife. The victim screams at 30 decibels. The stabbing pressure made by the knife and the victim's stomach is 10 Pa. The victim's height is 160 cm. The stabbed knife is 60 cm above the ground and is sticked at a depth of 5 cm into the victim's stomach. The length of the sharp part of the knife is 20 cm. Let S(t), P(t), H(t), and D(t) be decibels, amount of pressure, height of the stabbed knife above the ground, and the sticked depth of the stabbed knife into the victim's stomach at time t (in seconds). Every centimeter moved up the victim's body by the knife increases the decibel level, pressure, and sticked depth by 5 decibels, 7 Pa, and 3 cm respectively. Given that the psycho will move up the knife by k cm every second. Find:
a) the functions S(t), P(t), H(t), and D(t) in terms of k
b) the value of k so that the knife will have no sharp part left (completely sticks inside the victim's body) as it reaches the victim's forehead (full height)
c) Let X(t) be the psycho's satisfaction level at time t
given by:
X(t) = 8t³×S(t)+4t²×P(t)+t×H(t)+2D(t)
Calculate:
i) the time when the psycho will be the most
satisfied while having fun with the victim
ii) the highest satisfaction level the psycho can
experience
#Help with hard integration problem
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what book did you get this question from
I'd love to read it
S(t) = 30+5kt
P(t)=10+7kt
how does depth increases by 3cm every centimetre pushed in?
bro WHAT?
this is pretty easy
use Xenon's law of transformation
and youll get the answer
it's not from a book
my friend challenged me
he made it up himself
how?
its wrong
and you created a discord account just to ask this question that's why the username
explain this tho
no it means depth increases by 3 cm every centimetre moved up
I think so
this is write
every second knife moves up k cm
scream increases by 5k decible
D(t) = 5+ 3kt
H(t)=60+kt
I have an account previously but I deleted it, and today my friend challenged me so I created a new account to ask this
that's alright but do you understand my answer for the first question?
for part b
suppose knife reaches the forehead at T
60+kT=160
T=100/k
at the same time D(t)=20
5+3kT+20
T=5/k
so there's no value of k for which this is possible
ohh make sense
so the question is wrong?
$$X(t)=8t^3(30+5kt)+4t^2(10+7kt)+t(60+kt)+2(5+3kt)$$
$$X(t)=40kt^4+(240+28k)t^3+(40+k)t^2+(60+6k)t+10$$
for maximum satisfaction is at T where X'(T)=0 and its answers are too messy
scilent
no values is also a valid solution I think
,w maxima of 40kx^4+(240+28k)x^3+(40+k)x^2+(60+6k)x+10
oh wait
I'm an idiot
maximum satisfaction will be when knife is all the way in
so at t= 5/k
,w value of 40kx^4+(240+28k)x^3+(40+k)x^2+(60+6k)x+10 at x=5/k
@gleaming lynx ask you're friend if these answers are right
and also to see a therapist
why at t=5/k?
sure I'll ask him tomorrow
LMAO
because at t=5/k D(t)=20 which means knife is all the way in
,w plot of 40kx^4+(240+28k)x^3+(40+k)x^2+(60+6k)x+10
I personally think this is more correct
I think it is
for positive K maxima is at infinity it only gives answers for negative K
,w minima of 40kx^4+(240+28k)x^3+(40+k)x^2+(60+6k)x+10
,w minima of 40kx^4+(240+28k)x^3+(40+k)x^2+(60+6k)x+10 in domain (0, 5/k)
,w maxima of 40kx^4+(240+28k)x^3+(40+k)x^2+(60+6k)x+10 in domain (0, 5/k)
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yeah @gleaming lynx I think I am right
maximum satisfaction will be alwayse at t=5/k
I still don't get it
in normal case (k>0) its maxima is at infinity but the knife can't go infinitely in
I think I am wrong somewhere but I don't see it you'll have to ask your friend
ohh
okay then
I have another question
can I ask here or should I make a new post?
I thought the first question involves integration
2nd question:
An evil scientist is draining blood from a person's body at a rate of 10 ml/s into a cylindrical tank with a radius of 10 cm and a height of 20 cm. The tank is initially 1/10 full. Find:
a) the rate at which the height of the blood is increasing
b) the general equation of the height of the blood at time t (in seconds) h(t)
c) the general equation of the of volume of the blood at time t (in seconds) V(t)
also from my friend
bro really needs therapy
LOL
v= area * height
rate at which height is increasing:
$$100\pi h=10 cm^3$$
$$h=\frac{1}{10\pi}$$
scilent
$h(t)=2+\frac{t}{10\pi}$
scilent
$$V(t)=100\pi h(t)$$
$$V(t)=200\pi + 10t$$
scilent
10 ml blood is droping every second so area * (height rate) = volume rate
it did
$$V=AH$$
$$V'=AH'$$
scilent
x' is derivative of x
this part?
shouldn't the 2 be 20?
tank is 1/10 full so 20(1/10)=2
@gleaming lynx bro you get it?
ohh you're right
yes
@crystal pike thank you so much bro
appreciate it
@lavish temple has given 1 rep to @crystal pike
@crystal pike I've asked my friend and here is the modified version
A psycho is stabbing a person with a knife. The victim screams at 30 decibels. The stabbing pressure made by the knife and the victim's stomach is 10 Pa. The victim's height is 160 cm. The stabbed knife is 60 cm above the ground and is sticked at a depth of 5 cm into the victim's stomach. The length of the sharp part of the knife is 380 cm. Let S(t), P(t), H(t), and D(t) be decibels, amount of pressure, height of the stabbed knife above the ground, and the sticked depth of the stabbed knife into the victim's stomach at time t (in seconds). Every centimeter moved up the victim's body by the knife increases the decibel level, pressure, and sticked depth by 5 decibels, 7 Pa, and 3 cm respectively. Given that the psycho will move up the knife by k cm then drops it by 1 cm every second. Find:
a) the functions S(t), P(t), H(t), and D(t) in terms of k
b) the value of k so that the knife will have no sharp part left (completely sticks inside the victim's body) as it reaches the victim's forehead (full height)
c) Let X(t) be the psycho's satisfaction level at time t
given by:
X(t) = 8t³×S(t)+4t²×P(t)+t×H(t)+2D(t)
Calculate:
i) the time when the psycho will be the most
satisfied while having fun with the victim
ii) the highest satisfaction level the psycho can
experience
I think you can also do it with integration
for example S(t):
S(0) = 30
dS(t)/dt = 5k
dS(t) = 5k dt
then integrate both sides
S(t) = 5kt+C
plug in t=0
S(0) = 5k(0)+C
30 = C
C = 30
S(t) = 5kt+30
S(t) = 30+5kt
same for the other functions
I don't think you can call that a knife anymore
even "sword" seems like too less
LOL
idk I asked him and he just modified it that way