help with related rates @StevenNL2000

"A water tank in the shape of a right circular cone of radius 300 cm and height 500 cm leaks water from the
vertex at a rate of 10 cubic cm/minute. Find the rate at which the water level is decreasing when the height is
200 cm."
kinda a mean question but here you go. gonna assume you're not supposed to use integration so imma avoid it cause related rates usually dont use them
cone volume formula is ((pih)/3)r^2=V
we're solving for dh/dt at h=200 so first we gotta come up with a formula for height so we can compare it to time.
some algebra with the cone formula gives you (3V)/(pi*r^2) = h which is an issue cause r is dependent on h. using some basic similar right triangle laws we know that r=3h/5 some substitution and algebra later we get sqrt(5v/pi) = h which is much more workable. you should be able to do the calculus for the rest but if not lmk and ill go through it

its late at night and ill do the next question tomorrow

@"shrimp"#10 yes just implicit differentiation of each side with respect to t
so (d/dt)h and (d/dt)(sqrt(5v/pi))
= dh/dt = (sqrt(5))/ (2(sqrt((pi)(v))))(dv/dt)
now we just plug in conditions
v at h=200 = 1/3 pi 200 120^2
dv/dt = -10
still no calculator but at this point that's all you need

"A lighthouse light rotates at a rate of 3 revolutions per minute. The beam hits a wall located 10 miles away and
produces a dot of light that moves horizontally along the wall. How fast is this dot moving when the angle
between the beam and the line through the lighthouse perpendicular to the wall is pi/6?"

this question isnt really calculus as far as i can tell
3 rev/m = pi/10 rad/s
recall that angular speed is rad/s * radius
now we just make a triangle where the hypotenuse is the beam of light at pi/6 from perpendicular, the longer side is the perpendicular beam (10m) and the angle between the hypotenuse and the 10m side is pi/6

so to solve for the hypotenuse you do cos(pi/6)=10/h
cos(pi/6)= sqrt(3)/2 so we divide 10 by sqrt(3)/2 and that'll give us hypotenuse length. multiply by our pi/10 rad/s and that should be the speed at this distance. 2pi/sqrt(3) rad/s should be your answer.

keep posting these i need to keep my calc up to date

Damn, I'm taking calc and only half understand all these posts. Shits crazy man