Derivative and trigonometric functions

[Jhenrique]

Hellow!

If we can equal the first derivative with a trigonometric function:
\frac{dy}{dx}=tan(\theta)

So, the second derivative is equal to which trigonometric function?
\frac{d^2y}{dx^2}=?

Thanks!

 

[jedishrfu]

Hellow!

If we can equal the first derivative with a trigonometric function:
\frac{dy}{dx}=tan(\theta)

So, the second derivative is equal to which trigonometric function?
\frac{d^2y}{dx^2}=?

Thanks!

You need to fix your derivative: dy/dx = tan(x) would be correct.

then use tan(x) = sin(x) / cos(x) and the division rule.

 

[scurty]

It is easier to derive if you rewrite tangent in terms of sine and cosine and use the quotient rule.

$\frac{dy}{d\theta} = \frac{\sin{\theta}}{\cos{\theta}}$
$\frac{d^2y}{d\theta^2} = \displaystyle\frac{\cos{\theta}\cos{\theta} + \sin{\theta}\sin{\theta}}{\cos^2{\theta}}$

And then simplify.

 

[Mark44]

Hellow!

If we can equal the first derivative with a trigonometric function:
\frac{dy}{dx}=tan(\theta)

So, the second derivative is equal to which trigonometric function?
\frac{d^2y}{dx^2}=?

If I take what you wrote at face value, the y'' = 0, assuming that x and θ are independent of one another.

Assuming as others in this thread have done, that you meant dy/dx = tan(x), the problem can be simplified.

If w = tan(x), then dw/dx = ? The derivative of the tangent function is shown in every calculus book.

 

[Jhenrique]

\frac{dy}{dx}=tan(\theta)
proceeded from:
\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{\frac{dr}{d\theta}sin(\theta)+rcos(\theta)}{\frac{dr}{d\theta}cos(\theta)-rsin(\theta)}

But, I don't know how do d²y/dx²

 

[jedishrfu]

\frac{dy}{dx}=tan(\theta)
proceeded from:
\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{\frac{dr}{d\theta}sin(\theta)+rcos(\theta)}{\frac{dr}{d\theta}cos(\theta)-rsin(\theta)}

But, I don't know how do d²y/dx²

Can you state the whole problem? It appears that theta is a function of x and y.

 

[Jhenrique]

I just use the equality x = r cos(θ) and y = r sin(θ) to find that dy/dx is tan(θ) (it's known). But I don't know to manipulate all this differential algebra to find d²y/dx² as a trigonometric equation... Do you know how do it?

EDIT: In other words... my ask is if is possible to express d²y/dx² in term of theta

 

[jedishrfu]

I just use the equality x = r cos(θ) and y = r sin(θ) to find that dy/dx is tan(θ) (it's known). But I don't know to manipulate all this differential algebra to find d²y/dx² as a trigonometric equation... Do you know how do it?

EDIT: In other words... my ask is if is possible to express d²y/dx² in term of theta

I still don't see the problem you're trying to solve. The x=r cos(theta) and y=r sin(theta) are the conversion equations from polar to xy coordinate systems so what equation are you trying to differentiate.

 

[Mark44]

Then you should state this in your initial post. We assumed that when you wrote dy/dx = tan(θ) that you had written θ when you meant x.

I just use the equality x = r cos(θ) and y = r sin(θ) to find that dy/dx is tan(θ) (it's known).

I don't see how this can be true.

Since y = r sin(θ) and x = r cos(θ), then y/x = sin(θ)/cos(θ) = tan(θ). Of course, y/x and dy/dx are not the same thing.

But I don't know to manipulate all this differential algebra to find d²y/dx² as a trigonometric equation... Do you know how do it?

EDIT: In other words... my ask is if is possible to express d²y/dx² in term of theta

 

[Mark44]

\frac{dy}{dx}=tan(\theta)
proceeded from:
\frac{dy}{dx}=\frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}=\frac{\frac{dr}{d\theta}sin(\theta)+rcos(\theta)}{\frac{dr}{d\theta}cos(\theta)-rsin(\theta)}

The above doesn't make any sense to me. r is not a function of θ, so dr/dθ has no meaning. y is a function of θ, but it's also a function of r, so you would need to be dealing with partial derivatives. Same with x.

But, I don't know how do d²y/dx²

 

[Jhenrique]

It is in page 670 of book of James Stewart's Calculus (see the image archived)

I don't undestood very well, but, anyway, I think that is possible to express d²y/dx² through of theta too. Is possible?

 

[Mark44]

I don't read Portuguese, but I have a copy of Stewart's Calculus that contains the section you show. Stewart makes several assumptions that you didn't show - he assumes that θ is a parameter (i.e., essentially a constant) and that r is a function of θ. In other words, he writes r = f(θ). If you ask a question and don't include the assumptions that are being made, you're going to get answers like the ones you saw in this thread.

To find $\frac{d}{dx}(\frac{dy}{dx})$, differentiate your expression with respect to x. Using the chain rule, d/dx(g(θ)) = d/dθ(g(θ)) * dθ/dx. Here what I'm calling g(θ) is shorthand for the expression you have for dy/dx.

 

[jedishrfu]

I don't read Portuguese...

Darn, we could have made a movie and you'd be the next John Travolta as he was in 'Phenomena'
where he learned Portuguese on his way to rescue a missing boy.

 

[Jhenrique]

If you ask a question and don't include the assumptions that are being made, you're going to get answers like the ones you saw in this thread.

Sorry

Darn, we could have made a movie and you'd be the next John Travolta as he was in 'Phenomena'
where he learned Portuguese on his way to rescue a missing boy.

Hilarious!

 

[Mark44]

No hablo portuguesa ahora, pero quiero estudiar portuguesa alguna vez.