<p>Okay, I have a test tomorrow and I'm having trouble understanding how to do these problems. Here are a few examples. If you can, please show step by step how to do it and why you did each step. Thanks</p>
<p>1) The radius (r) of a right circular cone is increasing at a rate of 2 inches per minute. The height (h) of the cone is related to the radius by h = 3r. Find the rate of change of the volume when (A) r = 6 inches and (B) r = 24 inches.</p>
<p>2) A water tank in the form of an inverted cone is being emptied at the rate of 6 m^3/min. The altitude of the cone is 24 m, and the radius is 12 m. Find how fast the water level is lowering when the water is 10 m deep.</p>
<ol>
<li><p>V=(1/3)b<em>h=(1/3)(pi</em>r^2)(3r)=pi<em>r^3
dV=3</em>pi*r^2dr
dr=2; plug in r and dr to find dV</p></li>
<li><p>h=2r
V=(1/3)b*h=(2pi/3)r^3=(2pi/3)(h^3)/8
dV=(pi/4)h^2dh
plug in dV, h to find dh</p></li>
</ol>
<p>1/. we know dr/dt = 2in/min
h=3r
(a) r = 6 inches.
volume of a cone is V = (1/3) pi r^2h
but because h = 3r (always) --> V = (1/3)pi (r^2)(3r) = pi r^3
take the derivative of that --> dv/dt = 3pi r^2 (dr/dt)
substitute in the values: dv/dt = 3pi (6^2)(2) = 216pi (inches cube/min)
and that's the answer</p>
<p>Thanks...I have a few questions though...so whenever there is something changing... like in the first example the radius we put that in for "x", and the rate at which it changes we put in for dx/dt? Is this always the case? You always put the value in for the "x" and the rate in for the derivative? </p>
<p>What are the steps in solving these problems?</p>
<p>Do you first figure out the constants and variables? Then you setup the formula/equation and take the derivative? After that you plugin the values you have and solve for the missing variable?</p>
<p>Usually to solve related rates problem, what I do is:
1. List information that is known and information trying to find
2. Find equation to relate the derivatives
3. Find an equation that when differentiated will equal one of the derivatives you need from the above equation you created in step 2
4. Differentiate the equation from step 3
5. Use that derivative and the rate given in the problem to find the third rate of change</p>
<p>For instance, on my calc final today we had a related rates problem about the change of volume (at 40 ft) of an inverted cone with diameter = 200 ft and height=100 ft. The height was rising by 1 ft/ min. So I listed:</p>
<p>I know:
Radius= 100 ft
Height (of cone) = 100 ft
dH/dt = 1 ft min
v = 1/3(pi)(r^2)h</p>
<p>Trying to Find:
dV/dt at h=40</p>
<p>Next, I wrote an equation to relate this:
(dV/dt) = (dV/dH) * (dH/dt)</p>
<p>So, since I knew dH/dt, I knew that I would have to find dV/dH in order to find dV/dt. Since I was trying to find the rate of change of the volume with respect to height I tried to get an equation of v in terms of H.</p>
<p>v = 1/3(pi)(r^2)(h)
In this problem r was equal to h, so I didn't need to relate r and h.
Therefore V(h) = 1/3(pi)(h^2)(h) = 1/3(pi)(h^3)</p>
<p>Next, I differentiated that function to find dV/dh.
v'(h) = (3)(1/3)(pi)(h^2) = (pi)(h^2) = dV/dh</p>
<p>Then, you just have to use the relation we found above to find dV/dt.
dV/dt = dV/dh * dh/dt
dV/dt = pi (h^2) * 1 ft/ min
dV/dt at h=40: pi(40^2)= 1600pi feet/min</p>
<p>Anyway, I don't know if that helped at all...but usually it helps me to see examples worked out for calc instead of just reading about it in words and stuff.</p>
<p>I have no idea what to do here. I am begging for some help and I need it quick, here are some of the problems</p>
<p>1)Assuming that oil spills from a ruptured tanker spreads in a circular pattern whose radius at a constant rate of 2ft/ sec, how fast is the area of the spill increasing when the diameter of the spill is 120 Ft?</p>
<p>2)In order to record the launch of a rocket, a cameral is placed 3000 ft from the base of the lauching pad. If the rocket is rising vertically at a rate of 880 ft/sec, how fast is the camera-to-rocket distance when the rocket is 4000 ft vertically above the pad?</p>