Very very short question on second derivative

In summary: You first differentiate implicitly to get an equation relating y and y'. Then, you can use algebra to solve for y' in terms of y. Finally, you can substitute the given point (0, 3) to find the value of y' at that point.
  • #1
CookieSalesman
103
5
What does it mean when I have to find the second derivative of a circle at a given point? (Implicit diffing)

In specifics, the equation is 9x2 +y2 =9
At the point (0,3)
You don't really need the rest at all, but it was just my process.

This seems to make no sense.
first D'v 18x+2yy'=0
Second derivative gives 18+y''=0
Doesn't it?
How am I supposed to substitute coordinates?
 
Last edited:
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  • #2
CookieSalesman said:
What does it mean when I have to find the second derivative of a circle at a given point? (Implicit diffing)

In specifics, the equation is 9x2 +y2 =9
At the point (0,3)

This seems to make no sense.
first D'v 18x+2yy'=0
That equation is not the first derivative. You need to solve for y' algebraically.
CookieSalesman said:
Second derivative gives 18+y''=0
Doesn't it?
No. When you have isolated y', differentiate it to get y''. Then, to find the value of y'' at (0, 3), substitute these coordinates in your formula.
CookieSalesman said:
How am I supposed to substitute coordinates?
 
  • #3
CookieSalesman said:
What does it mean when I have to find the second derivative of a circle at a given point? (Implicit diffing)
Strictly speaking, it doesn't mean anything! You do not take the derivative of geometric objects, you take the derivative of functions. What you are asking about the relation describing the circle.
In specifics, the equation is 9x2 +y2 =9
At the point (0,3)
That's not a circle, it's an ellipse.
This seems to make no sense.
first D'v 18x+2yy'=0
Second derivative gives 18+y''=0
Doesn't it?
How am I supposed to substitute coordinates?
Using "implicit differentiation" as you say, you get 18x+ 2yy'= 0 where y' is the derivative with respect to x.
Doing that again you do NOT get 18+ y''= 0. You forgot to use the product rule on 2yy'. Implicit differentiation gives, rather,
 
Last edited by a moderator:
  • #4
Okay, so if I differentiate once implicitly, I get 18+2yy'
Then, algebra, so that it's -9/y = y'
But if I differentiate that, then I have y prime and y double prime...
Oh, so I can find y prime from the earlier equation, then resubsitute?

Thanks.
 
  • #5
CookieSalesman said:
Okay, so if I differentiate once implicitly, I get 18+2yy'
Then, algebra, so that it's -9/y = y'
But if I differentiate that, then I have y prime and y double prime...
Oh, so I can find y prime from the earlier equation, then resubsitute?

Thanks.
That's the idea.
 

1. What is a second derivative?

A second derivative refers to the rate of change of the slope of a function. It is the derivative of the derivative, and it measures how quickly the slope of a function is changing at a specific point.

2. How is the second derivative calculated?

The second derivative is calculated by taking the derivative of the first derivative. This can be done using the power rule, product rule, quotient rule, or chain rule depending on the function.

3. What does a positive second derivative indicate?

A positive second derivative indicates that the slope of the function is increasing, meaning the function is concave up. This can also mean that the function is gaining speed or accelerating.

4. What does a negative second derivative indicate?

A negative second derivative indicates that the slope of the function is decreasing, meaning the function is concave down. This can also mean that the function is losing speed or decelerating.

5. How is the second derivative used in real-world applications?

The second derivative is often used in physics and engineering to describe the acceleration of an object in motion. It can also be used in economics to analyze the rate of change of a function, such as the marginal cost or marginal revenue of a business.

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