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[frostshotgg]

Specific heat of water = 4.186 J/g. deltaT from 22C and 37C = 15C. 15*4.186 = 62.79 J/g to get to 37. deltaT from 55 to 37 = 18C. 18*4.186=75.348 J/g of heat to give up. 62.79*90 = 5651.1J to heat up the water. 5651.1/75.348 = 75g of 55C water.

[Powder Miner]

What would the parent formula be for that?

[frostshotgg]

I guess the specific heat equation? delta = Q/mc.

[Powder Miner]

No, I'm running low on sleep and I'm being really really dumb-- the formula for calculating the amount of water needed using specific heat.

[frostshotgg]

F = final temperature (In the example you gave 37). T1 = low temperature (In the example you gave 22). T2 = high temperature (In the example you gave 55). C = 4.186. m1 = mass of cool water (90g) m2 = mass of hot water (answer, 75g).

(((F-T1)*C)*m)/((T2-F)*C) = m2

EDIT: Worth noting, C cancels out in this case, although if it were two different liquids it would matter. In your case, (F-t1)*m1 / (t2-f) = m2.

[Powder Miner]

Thank you!

[Gatleos]

I guess I might as well post this here rather than in the programming thread.

Say I have a vector expressed in x and y coordinates, and I want to reduce its length along some arbitrary axis by a certain fraction t, expressed as a number between [0.0, 1.0]. So t = 0.0 would set its length along the axis to zero, and t = 1.0 would leave it as-is.

For example, if the vector was (3, 2) and you set the axis you want to change its magnitude on to the y axis, and the fraction to t = 0.5, you would end up with (3, 1). And t = 0.0 would give you (3, 0), etc. This is easy enough with an orthogonal axis, since in that situation all you need to do is multiply t by one component of the vector. But on an arbitrary axis I can't seem to get the math to work.

I'm terrible at trigonometry help

[Gatleos]

Thank you, that was very helpful. Though I ended up multiplying the rejection by t, not the projection.

[Spehss _]

The derivative of e^(anything) is just e^(anything) right? And the derivative of that derivative would also be e^(anything)?

Like f(x)=e^(-1/(x+1)), then f'(x)=e^(-1/(x+1)), and f''(x)=e^(-1/(x+1)), right?

[da_nang]

The derivative of e^x is e^x. The derivative of e^f(x) is f'(x)e^f(x).

[Graknorke]

The derivative of e^(anything) is just e^(anything) right?

Nope. if f(x)=e^x then f'(x)=e^x, but if your function is of a higher order then you'll need to apply the chain rule.

[Spehss _]

The derivative of e^(anything) is just e^(anything) right?

Nope. if f(x)=e^x then f'(x)=e^x, but if your function is of a higher order then you'll need to apply the chain rule.

uuuuuuugh more work

[Spehss _]

Double post, sorry. no one else has posted, and I have a question.

Wanna make sure I have this right. The derivative of 5^(x) is 5x^(x-1), right? Because of the power rule? So assuming c is any constant, then derivative of c^(x) is cx^(x-1).

But my notes I have written down say that the derivative of c^(x) is c^(x)*lnx

[Reelya]

The power rule only works for constant powers. Forget about it when the power is a variable.

You have to apply the change of base rule. Let me work through this since I'm rusty on maths too and it helps me remember stuff:

Say you have y = c^x and you want this in ln, i.e. base e:

Spoiler: steps (click to show/hide)

First, take the natural log of both sides:
ln y = ln (c^x)

Then use the powers rule for logs:
ln y = x(ln c)

And then reverse the ln operation:
y = e^{x(ln c)}

Therefore, c^x = e^{x(ln c)}

Now, you want to take the derivative of e^{x(ln c)}, and you use the chain rule for that as above.

[Reelya]

It's been years since I did calc, and I'm not sure what rule you did in step 3 there to get dy/y = ln(c)dx ?