All categories

3 years ago

Is Time a function of temperature?

Mathematics

1 answer:

Naya [18.7K]3 years ago

6 0

Answer:

In general, time is not a function of temperature, but this could depend on the problem you're solving. For example, imagine a ball of metal cooling down, and data was given on how long the ball was cooling down and the temperature at the center. Someone could have fit a curve or function to the temperature to show how much time has elapsed when inputting temperature as the independent variable.

Step-by-step explanation:

You might be interested in

Simplified what is 5(x+1)

Anuta_ua [19.1K]

5(x + 1)....distribute the 5 thru the parenthesis...that means multiply the 5 by everything in the parenthesis.

5(x + 1) = 5 * x + 5 * 1 = 5x + 5

7 0

2 years ago

Read 2 more answers

What is the formula you use to measure the entire length of a diagonal

zheka24 [161]

Answer:

Pythagorean theorem

Step-by-step explanation

a^2+b^2=c^2

8 0

2 years ago

Use the Chain Rule (Calculus 2)

atroni [7]

1. By the chain rule,

$\dfrac{\mathrm dz}{\mathrm dt}=\dfrac{\partial z}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial z}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}$

I'm going to switch up the notation to save space, so for example, $z_x$ is shorthand for $\frac{\partial z}{\partial x}$ .

$z_t=z_xx_t+z_yy_t$

We have

$x=e^{-t}\implies x_t=-e^{-t}$

$y=e^t\implies y_t=e^t$

$z=\tan(xy)\implies\begin{cases}z_x=y\sec^2(xy)=e^t\sec^2(1)\\z_y=x\sec^2(xy)=e^{-t}\sec^2(1)\end{cases}$

$\implies z_t=e^t\sec^2(1)(-e^{-t})+e^{-t}\sec^2(1)e^t=0$

Similarly,

$w_t=w_xx_t+w_yy_t+w_zz_t$

where

$x=\cosh^2t\implies x_t=2\cosh t\sinh t$

$y=\sinh^2t\implies y_t=2\cosh t\sinh t$

$z=t\implies z_t=1$

To capture all the partial derivatives of $w$ , compute its gradient:

$\nabla w=\langle w_x,w_y,w_z\rangle=\dfrac{\langle1,-1,1\rangle}{\sqrt{1-(x-y+z)^2}}}=\dfrac{\langle1,-1,1\rangle}{\sqrt{-2t-t^2}}$

$\implies w_t=\dfrac1{\sqrt{-2t-t^2}}$

2. The problem is asking for $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ . But $z$ is already a function of $x,y$ , so the chain rule isn't needed here. I suspect it's supposed to say "find $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}$ " instead.