# Thread: Using calculus

1. ## Using calculus

Hi there,

I am struggling with this problem I have to use calculus to find an expression for the rate of change of temperature with respect to time;

T = 25 - 4cos((π(t-3))/12)

This is the equation in which I need to differentiate. I haven't differentiated sine or cosine functions before?

*note that π is supposed to be pi = 3.142

If anybody could help me, your help is mostly appreciated

2. Originally Posted by Mercy99
Hi there,

I am struggling with this problem I have to use calculus to find an expression for the rate of change of temperature with respect to time;

T = 25 - 4cos((π(t-3))/12)

This is the equation in which I need to differentiate. I haven't differentiated sine or cosine functions before?

*note that π is supposed to be pi = 3.142

If anybody could help me, your help is mostly appreciated
First of all, $\frac{d}{dx}\cos(x)=-\sin(x)$. This can be discovered graphically, or through the definition of the derivative.

When you differentiate, you need to apply the chain rule:

$T=25-4\cos\bigg(\frac{(t-3)\pi}{12}\bigg)=25-4\cos\bigg(\frac{\pi t}{12}-\frac{3\pi}{12}\bigg)$

Thus,

$\frac{dT}{dt}=4\sin\bigg(\frac{\pi t}{12}-\frac{3\pi}{12}\bigg)\cdot\frac{\pi}{12}=\color{re d}\boxed{\frac{\pi}{3}\sin\bigg(\frac{\pi t}{12}-\frac{3\pi}{12}\bigg)}$

Does this make sense?

--Chris

3. Two questions;

Where did the 25 - part go? and where did the π/12 part come from?

Other than that, that was a really good help, thank you!

4. Originally Posted by Mercy99
Two questions;

Where did the 25 - part go? and where did the π/12 part come from?

Other than that, that was a really good help, thank you!
The derivative of a constant is zero. Here's a quick proof using the definition of the derivative:

If $f(x)=c$, where c is a constant, then by the definition of the derivative, $f'(x)=\lim_{h\to{0}} \frac{f(x+h)-f(x)}{h}=\lim_{h\to{0}}\frac{c-c}{h}=\lim_{h\to{0}}0=\color{red}\boxed{0}$

This is where $\frac{\pi}{12}$ came from:

$\frac{dT}{dt}=4\sin\bigg(\frac{\pi t}{12}-\frac{3\pi}{12}\bigg)\cdot \frac{d}{dt}\bigg[\frac{\pi t}{12}-\frac{3\pi}{12}\bigg]$

The derivative of $\frac{\pi}{12}t-\frac{3\pi}{12}$ is $\frac{\pi}{12}$

Does this clarify things?

--Chris

5. For that equation, the y-value is the same (it is always -0.5236)
When typed into a graphics calculator, the equation is just a constant line?
In the question, it says that 'whenever the rate of change of temperature, with respect to time, is greater than or equal to +0.2(degrees Celsius) per hour, the system switches on. It switches off again when the rate of change of temperature, with respect to time is less than +0.2(degrees Celsius) per hour.

Hence find the range of values of t (correct to two decimal places) for which the watering system will be on.

Shouldn't the derivative be oscillating?

6. Originally Posted by Mercy99
For that equation, the y-value is the same (it is always -0.5236)
When typed into a graphics calculator, the equation is just a constant line?
In the question, it says that 'whenever the rate of change of temperature, with respect to time, is greater than or equal to +0.2(degrees Celsius) per hour, the system switches on. It switches off again when the rate of change of temperature, with respect to time is less than +0.2(degrees Celsius) per hour.

Hence find the range of values of t (correct to two decimal places) for which the watering system will be on.

Shouldn't the derivative be oscillating?
It does.

Originally Posted by Chris L T521

$\frac{dT}{dt}=4\sin\bigg(\frac{\pi t}{12}-\frac{3\pi}{12}\bigg)\cdot\frac{\pi}{12}=\color{re d}\boxed{\frac{\pi}{3}\sin\bigg(\frac{\pi t}{12}-\frac{3\pi}{12}\bigg)}$
The sine function is an oscillating function...

--Chris