Anyone able to help me out with calculus homework?

Gyorgy porgy ready for an orgy.

If the slav is walking away from the light source, then the head of its shadow should be moving away from it, so yeah you're right

But the problem, although flawed, is still solvable, correct?

ok if you change it to the shadow is moving 25 times his speed (so speed*25 instead of speed*0.25) you can do it.
25 times the speed means the circumpherance of the light area coming off the lamp is 25 times larger than the circumpherace of the path your taking around the lamp. and you must be taking a circular path around the lamp or the question is too ambiguous to answer correctly.
assume your path is radius=1 unit and we know the lamp is 5 units high so we have a formula involving lamp height, length and height and we only know lamp height and you can't solve for two variables so your fucked.

this question is so fucking stupid.

the author probably is trying to say the shadow is GROWING at a rate of 0.25*your speed, which would also be stupid because you still need shadow length to get it down to one variable to solve for.

tldr, no your teacher was fucking high when he wrote that question.

Just a FYI, the teacher only remains at the school because he is entirely invested in the drama department.

I dunno its been ages since i did calc

If the situation was strictly two dimensional on a right triangle, how would you go about the problem?

Two remaining problems. 1/2

2/2 I hate my life

that is the situation, actually. the circle bullshit is just a way to get the ratio between the location of the height line on the bottom side of the triangle relative to the location of the right point of the triangle.
even thats not enough though because you need 2 variables to get the 3rd side of the triangle or height of the person or the shadow length, get any one of these values and you can solve for the rest. but you don't have them and they could be anything.

car is moving at the same speed as the guy pulling the rope. the pully part is just a red herring and is irrelevant to the problem which is basically "if you pull a rope at x speed how fast does the other end move?"
would be a different story if you had more than one pully and rope which could be geared to change end relative motion of the ropes but with one pully and one rope its irrelevant.

cant read that shit.

A spherical iron ball 8 inches in diameter is covered with a layer of ice of uniform thickness. The ice is melting at a constant rate of 10 in^3/min.

a) How fast is the thickness of the ice decreasing when it is 3 inches thick?

b) How fast is the outer surface area of the ice decreasing when the ice is 2 inches thick?

Draw a line from head to lamppost (orthogonal to lamp post). Call that x, call line from head to top of lamp D and line from head to shadow L.
At time 0 man and shade are both inside the lamppost.
Angle between D and the lamppost is p.
x is position of man, s is of shade.

h=L*cos(p)
5=(L+D)*cos(p)
L=h/cos(p)
L+D=5/cos(p)
x=D*sin(p)=h*tan(p)
s=(L+D)*sin(p)=5*tan(p)
dx/dt=4*ds/dt
=h*(tan(p))'=4*5*(tan(p))'
h=20

*Guyorgy

Assuming the rope doesn't stretch, the car will move at the same speed as the rope.
Hence 2.5 feet/second

>man is walking away from lamp post
>away
>shadow moving slower
shadow would be GROWING (not moving) faster and faster as he walked AWAY from lamp.

ok you got two variables: sphere with radius 4in and ice decreasing 1 cubic inch per minute.
>How fast is the thickness of the ice decreasing when it is 3 inches thick?
ice is melting at a constant rate so the >when it is 3 inches thick?
is another red herring, it is irrelevant, the answer is always 10in^3/min until the ice is gone, then its 0.
>
b) How fast is the outer surface area of the ice decreasing when the ice is 2 inches thick?
see above

the size of the iron ball is also a red herring. its irrelevant to the questions.

this is stupid how is that supposed to test for any knowledge except reading comprehension?