Very strange question for derivative

  • Thread starter Thread starter peacemaster
  • Start date Start date
  • Tags Tags
    Derivative Strange
peacemaster
Messages
14
Reaction score
0

Homework Statement



The question in it's complete wording:

Consider the following curve.
y = 8 + 5x2 - 2x3.

Find the slope of the tangent to the curve at the point where x = a.

Homework Equations



(y2-y1)/(x2-x1) where x2=x1+dx and the limit of dx is approaching zero and y2=f(x+dx).

The Attempt at a Solution



Well, this is the really hard part because they don't tell you what a is. Basically, this is a curve and they say find the slope of the tangent at the point where x=a, but that point could be anywhere on the entire curve.

Don't worry, I didn't misread the question. There isn't any info lingering elsewhere. A is not defined anywhere in the problem.

So my question is... what gives? Am I missing something here?

Thank you so much for any help,

Peacemaster
 
Last edited:
Physics news on Phys.org
peacemaster said:

Homework Statement



The question in it's complete wording:

Consider the following curve.
y = 8 + 5x2 - 2x3.
Better notation: y= 8+ 5x^2- 2x^3. Even better is to use "LaTex": y= 8+ 5x^2- 2x^3

Find the slope of the tangent to the curve at the point where x = a.


Homework Equations



(y2-y1)/(x2-x1) where x2=x1+dx and the limit of dx is approaching zero and y2=f(x+dx).

The Attempt at a Solution



Well, this is the really hard part because they don't tell you what a is. Basically, this is a curve and they say find the slope of the tangent at point a, but point a could be anywhere on the entire curve.

Don't worry, I didn't misread the question. There isn't any info lingering elsewhere. A is not defined anywhere in the problem.

So my question is... what gives? Am I missing something here?

Thank you so much for any help,

Peacemaster
You have written two contradictory things. You say "at the point where x = a" and then, later, "find the slope of the tangent at point a". Is "a" a number or a point?

If a is a number, "x= a", then I don't see what the difficulty is. If the problem had said "at the point where x= 0" or "at the point where x= 2" what would you have done? Whatever you do with "0" or "2", do the same with "a".
 
Wow, good catch. I didn't realize the contradiction. The way I stated the problem is all the info given to me by the question.

The problem asks to find the slope of the tangent at the point where x=a. A is an x value, not a point. The problem does not tell me what a is.

I will edit the original post to reflect the clarification.

What would you do with the problem the way I originally posted it?

Thanks,

Peacemaster
 
peacemaster said:
Wow, good catch. I didn't realize the contradiction. The way I stated the problem is all the info given to me by the question.

The problem asks to find the slope of the tangent at the point where x=a. A is an x value, not a point. The problem does not tell me what a is.

I will edit the original post to reflect the clarification.

What would you do with the problem the way I originally posted it?

Thanks,

Peacemaster

You can treat a as any number. As HallsofIvy has already pointed out.

HallsofIvy said:
If the problem had said "at the point where x= 0" or "at the point where x= 2" what would you have done? Whatever you do with "0" or "2", do the same with "a".

a can be 0, 2, 3, or 1/3, blah blah blah. It does not matter, you just think of a as some number that you don't know, or maybe don't need to know.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top