Callum

I've been asked to find the second derivative of the following formula. This is not simply me asking you to do it, I have spent hours attempting this. I have the answer but can't fully understand how to get there.

The formula is.

y^3 + x + sin(y) = 1

Now I have found the first derivative the following way.

3y^2*dy/dx + 1 + cos(y)*dy/dx = 0

Rearranged to

dy/dx = - ( 1 / (3y^2 + cos(y) ) )

This wasn't difficult and it's the correct answer. However getting the second derivative has left me stumped.

I'm not sure how to approach it, and any help would be amazing!

The answer is;

( sin(y) - 6y ) / ( (3y^2 + cos(y) )^2 )

But I can't work out the process.

I can probably calculate this myself without to much trouble, however I thought I'd share this with you all.

Therefore, all I need to do to get out of a speeding ticket is show that the light was green to me. Haha

What are some methods of calculating the following values, that would have been available in Newton's time?

The radius of the Earth,

The value of the acceleration due to gravity on Earth's surface,

The orbital distance of the moon,

and the orbital period of the moon.

10 Points for the best answer!

I'm having trouble with a question involving flow rate. Here is the question.

There is a wall holding back the ocean, and 2 metres below sea level there is a small hole in the wall with a diameter of 1.2cm.

Starting from the moment the hole appeared, how long would it take the water pouring out to fill a 1km^2 area to the height of 1m?

The answer is given as 1.76x10^6 seconds, however my answer doesn't match.

Using Bernoulli's equation, and assuming that the speed that the sea level is decreasing is 0, I calculated that the speed of the liquid should be sqrt(2gh) which would be 6.3m/s

And then the flow rate should be velocity * Area of the hole.

But when I then divide 1,000,000m^3 by the flow rate, I get 1.4x10^9s

WHERE AM I GOING WRONG!!