Partial Derivate of multivariable cos

[Adyssa]

Homework Statement

Partial derivative of R = vt cos x with respect to t

Homework Equations

NA

The Attempt at a Solution

Keeping the other variables constant, I get 'v cos x'. (t is 1)

I just want to check this, because it's part of a larger problem and I feel my answer is way off because of this function. I'm a bit rusty on my calculus, it's been ... 13 years since I did it in high school!

Also, if I calculate the partial derivative with respect to x, I get '-vt sin x'?

 

[dynamicsolo]

If v and x are functions of t, then vt cos(x) is actually a product of three terms. You will need to differentiate this using the Product Rule.

 

[Adyssa]

OK, I think I'm confused as to what a Partial Derivative is then. According to the wiki page example:

[URL]http://upload.wikimedia.org/math/c/e/b/cebdc47218ace0b076c636d3e073f2c5.png[/URL]

[URL]http://upload.wikimedia.org/math/b/0/b/b0b10dc199691e38d0c9ccf8d7aec19a.png[/URL]

In the above example the xy term becomes y, whereas if the product rule was used, it would be x'y + y'x = y + x, or have I missed something?

Let me elaborate on the problem I'm trying to solve, it's an error bound and the equation is:

|Error(R)| <= |dR/dv(v)||Error(v)| + |dR/dt(t)||Error(t)| + |dR/dx(x)||Error(x)|

where dR/dv(v) is the Partial Derivative of R = vt cos x wrt v (multiplied by the error in v), and similarly for the two other derivatives wrt t and x.

 

[vela]

I take it x is the angle a projectile is being launched at, so it's not a function of time. Usually, the angle is denoted by a Greek letter, like θ, and the displacement in the horizontal direction, which is a function of time, is denoted by x.

So as far as your original post goes, your partial derivatives are correct.

 

[Adyssa]

Oh, yes it is an angle (in radians) and the symbol was phi, but I didn't have it at hand, so I just used x. I didn't realize that would alter the equation. Thanks for the clarity! :)

 

[Fredrik]

OK, I think I'm confused as to what a Partial Derivative is then. According to the wiki page example:

[URL]http://upload.wikimedia.org/math/c/e/b/cebdc47218ace0b076c636d3e073f2c5.png[/URL]

[URL]http://upload.wikimedia.org/math/b/0/b/b0b10dc199691e38d0c9ccf8d7aec19a.png[/URL]

In the above example the xy term becomes y, whereas if the product rule was used, it would be x'y + y'x = y + x, or have I missed something?

You have missed that the derivative of y with respect to x is 0. So one of the terms you get when you use the product rule on "xy" is 0. By the way, you shouldn't write (xy)'=x'y+xy', because this notation doesn't reveal what functions you're taking the derivative of. I like the notation $$\frac{d}{dx}(xy)=\bigg(\frac{d}{dx}x\bigg)y+x\bigg(\frac{d}{dx}y\bigg)=1y+x0=y.$$ Alternatively, you can define two functions f and g by $f(x)=x$ for all x, and $g(x)=y$ for all x. (Note that this makes g a constant function). Then you can write (fg)'(x)=f'(x)g(x)+f(x)g'(x)=1y+x0=y.

Regarding what a partial derivative is...don't think of it as something different from an ordinary derivative. It isn't. If you're asked to compute the partial derivative of xy² with respect to x, it means this: Let f be the function defined by f(t)=ty² for all t. Find f'(x) (i.e. the derivative of f, evaluated at x). If you're asked to compute the partial derivative of xy² with respect to y, it means this: Let g be the function defined by g(t)=xt² for all t. Find g'(y) (i.e. the derivative of g, evaluated at y).

The partial derivative of vt cos x with respect to t...that's the ordinary derivative of the function that takes t to vt cos x (as opposed to e.g. the function that takes x to vt cos x).

See what I mean when I say that a partial derivative isn't really something different from an ordinary derivative? The "with respect to" part of it is just telling you which function to take an ordinary derivative of.

 

[Adyssa]

Thanks Fredrik, that clears things up a lot, I appreciate the thorough explanation.