# Thread: Implicit Diff Story probs

1. ## Implicit Diff Story probs

Sorry if this causes any trouble but could someone check my answers for these problems?: 1. A point is moving along the graph of the function so that dx/dt is 2 cm/s. Find dy/dt for the indicated values of x. The function is y=sinx. The values are a)x= Pi/6 b)x= pi/4 c) x=pi/3 and d) x=pi/2. After finding these values tind the rate of change of the distanse between the origin and the moving point on the graph of the function. (x and y are functions of time).

I got a) the square root of 3 b) the square root of 2 c) 1 and d) 0.
for the rate of change part i had some ridiculous numbers and was lost. I got 2pie/(the square root of (pi squared +4)).

The second problemis: find the rate of change of the volume of a cone if dr/dt is 2 inches/min and h=3r when a) r=6 inches and b)r=24 inches. I got a) 216*pie and b) 576*pie

The third problem is: A conical tank (with vertex down), is 10 feet across the top and 12 ft deep. if water is flowing into the tank at the rate of 10 cubic feet per minute, find the rate of change f the depth of the water the instant it is 8 ft deep. I got 2.7/pi in/min for the answer.

The fourth problem is that a trough is 12 ft long and 3 ft across the top. its ends are isosceles triangles with an altitude of 3 ft. water is being pumped into the trough at 2 cubic feet per minute. How fast is the water level rising when it is 1 foot deep? i got 2 ft/min

The fifth and final problem is: a construction worker pulls a 16-foot plank up the side of a building under construcion by means of a rope tied to the end of the plank. assume the opposite end of the plank follows a path perpendicular to the wall of the building and the worker pulls the rope at the rate of .5 ft/s. how fast is the end of the plank sliding along the ground when it is 8 ft from the wall of the building. i got .5 ft/s for that one

Thanks for any help at all.

2. Pie?. Come on!

3. hahahahah i didnt even notice i put that ha thats funny thanks

4. For the conical tank problem, use similar triangles.

$\frac{r}{h}=\frac{5}{12}$

$r=\frac{5h}{12}$

Using the volume of a cone formula:

$V=\frac{1}{3}{\pi}(\frac{5h}{12})^{2}h=\frac{25{\p i}h^{3}}{432}$

Now, differentiate wrt time:

$\frac{dV}{dt}=\frac{25{\pi}h^{2}}{144}\frac{dh}{dt }$

Now, plug in the knowns and solve for dh/dt:

$10=\frac{25{\pi}(8)^{2}}{144}\frac{dh}{dt}$

$\frac{dh}{dt}=\frac{9}{10\pi}\approx{0.287} \;\ ft/min$

or 3.44 in/min

5. thanks. i used similar triangles but found out what r was when h is 8. I wonder why the answer wasn't the same? Your way is right because i did some problems out of my book and the answers checked out all right for those. I did the trough problem your way and got 4/3 ft/min. That is correct is it not? For the ladder problem i got -square root of 192 divided by 16, correct also? Please reply....thanks

6. For the ladder problem, you are.....................correct!

For the trough problem, do you mean the slant sides are 3 feet or the height of the trough is 3 feet?. The statement could go both ways.

8. the altitude slash height is 3. the isosceles sides are not -sorry i did not see the trough message

9. Since the height of the trough is 3 and the top is 3, we can do similar triangles again.

$\frac{b}{h}=\frac{1}{2}$

$b=\frac{h}{2}$

The volume of the trough is $V=12bh$

$V=12\frac{h^{2}}{2}=6h^{2}$

Now, differentiate:

$\frac{dV}{dt}=12h\frac{dh}{dt}$

We want dh/dt when h=1:

$2=12(1)\frac{dh}{dt}$

$\boxed{\frac{dh}{dt}=\frac{1}{6}} \;\ ft/min$

10. wouldn't it be 1? if we did similiar triangles it would be 3/3 and 1/unknown which is 1? The volume of a trough is v=1/2bh^2 correct? since b=h i substitued h in for be giving v=1/2h^3. the deriv. of that is dv/dt= 3/2h^2*dh/dt. h=1 leaving 2=3/2*dh/dt meaning dh/dt=4/3?
Edit: my bad isnt the area for a trough v=1/2b*h*l? or is your formula correct? -dang that was another stupid mistake i made
Edit2: how can the volume be 12bh when the length can differ?

11. nvm i see.

12. Originally Posted by jarny
nvm i see.