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Math Help - Using calculus

  1. #1
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    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
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mercy99 View Post
    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
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  3. #3
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    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!
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mercy99 View Post
    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
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  5. #5
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    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?
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mercy99 View Post
    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.

    Quote Originally Posted by Chris L T521 View Post

    \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
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