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I'm very confuzzled as to where the variable is here. I don't know where to differentiate.

Here's the question, differentiate:

$$y = \frac{\sin(\theta)}{2} + \frac{c}{\theta}$$

Do I solve for $f(c)$ or $f(\theta)$? I tried solving it for $f(c)$ treating theta as a constant, but I'm unsure if that's correct.

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    $\begingroup$ It's normally the other way around. Diferentiate with respect to theta keeping c as constant. Besides, the question would be useless if it were in terms of c, having a huge constant like $sin(theta)/2$ wouldn't make sense other than to confuse! $\endgroup$ Apr 3, 2016 at 5:22
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    $\begingroup$ 'a' for area, 'b' for breadth, 'c' for constant... $\endgroup$
    – manshu
    Apr 3, 2016 at 7:15
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    $\begingroup$ In any case, it would be a good idea to explicitly state your assumption. Something like "assuming with respect to θ" before your answer. $\endgroup$
    – BenM
    Apr 3, 2016 at 20:46

6 Answers 6

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Standard or common notation would have $c$ be the constant, and $\theta$ the variable.

In Calc I, early letters ($a, b, c, d \ldots$) are ususally constants, and late letters ($x, y, z$) are usually variables. $\theta$ is often used for a variable when the variable in question is an angle, just as $t$ is used for time, or $r$ for a radial distance.

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In strict terms, this is ambiguous. This may be $y(\theta)$ or $y(c)$; in each case the other variable is fixed. However, as noted, the likely intended interpretation is that $\theta$ is the variable.

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    $\begingroup$ One can also differentiate with respect to any $t$ to get simply $dy/dt = 0$... (Of course, assuming $\theta$ and $c$ being independent on $t$.) $\endgroup$
    – CiaPan
    Apr 4, 2016 at 11:24
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The correct answer is:

O Questioner, differentiate with respect to what???

Do you want $\frac{dy}{dc}$ or $\frac{dy}{dθ}$? Or perhaps $\frac{dθ}{dc}$?

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  • $\begingroup$ @Deusovi: Sorry I made a mistake in my previous comment. $\frac{dc}{d2}$ is undefined because $Δ2 = 0$ no matter what, so $\frac{Δc}{Δ2}$ is undefined at any point and hence $\frac{dc}{d2}$ is not either. Well it's sad when people are taught to differentiate but it's not specified clearly what they differentiate with respect to.. It's as if they don't even know the goal of finding a derivative... $\endgroup$
    – user21820
    Apr 5, 2016 at 1:03
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As x and y are to lengths, heights, quantities, etc., θ and φ are to angles. Moreover, c generally represents some constant.

This is particularly true in physics, where θ often refers to an angle or phase, and c represents the (constant) speed of light.

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It's a convention to represent constants by the letters c,d,a etc. $\theta$ can be treated as a variable intuitively (especially in Physics). Differentiation is represented in the form $\displaystyle\frac{d}{dx}$ or $\displaystyle\frac{d}{dy}$ or $\displaystyle\frac{d}{dz}$ of something. Here the letter $x,y$ or $z$ in the denominator (denominator is not actually correct as this is not a fraction; but we use it for easing up things) represents the variable.

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I believe that the intent of the exercise is to differentiate with respect to $\theta$, but you should keep in mind that you may also have to differentiate the function with respect to $c$, or even differentiate with respect to NONE of the variables shown, in which case the value would simply be $0$.

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