The ellipse x^2+3y^3=13, has two points when x+-1
. Find the slope of the ellipse at both these points.
2. Thoroughly explain the method you used to find the slope in part 1. What are the strengths
and weaknesses of the method you used?
3. Explain whether your method will work to find the slope of x^2+xy+y^2=3 at points -1, 2 and
(-1, -1 . If it succeeds, please show your work as you find these two slopes. If your method fails,
that’s fine, just explain thoroughly why it failed.
I will assume you meant
x = -1 , (the + and + are on the same key)
x^2 + 3y^2 = 13
2x + 6y dy/dx = 0
dy/dx = -2x/6y = -x/3y
when x = -1 ......
1 + 3y^2 = 13
3y^2 = 12
y^2 = 4
y = ± 2
the two points are (-1,2) and (-1, -2)
at (-1,2) , slope = 1/6
at (-1,-2) , slope = 1/-6 = -1/6
2. your call
3. x^2 + xy + y^2 = 3
2x + xdy/dx + y + 2ydy/dx = 0
dy/dx(x + 2y) = -2x - y
dy/dx = (-2x - y)/(x+2y)
at (-1,2) , slope= (2-2)/(-1+4) = 0
at (-1,-1) , slope = (2+1)/(-1-2) = -1
I don't get the instructions.
" If your method fails, that’s fine, just explain thoroughly why it failed." ---- > very odd!
Thank you! helped a lot and yes I meant '=' ,
basically I think he just wants us to explain on why it did or didn't fail, yes I know very strange! :)
1. To find the slope of the ellipse at the points x = 1 and x = -1, we need to take the derivative of the equation with respect to y and then solve for dy/dx.
Given equation: x^2 + 3y^3 = 13
First, let's take the derivative with respect to y:
d/dy (x^2) + d/dy (3y^3) = d/dy (13)
0 + 9y^2 dy/dx = 0
Next, let's solve for dy/dx:
dy/dx = 0 / (9y^2)
dy/dx = 0
Therefore, the slope of the ellipse at both x = 1 and x = -1 is 0.
2. The method I used to find the slope in part 1 is called implicit differentiation. This method is useful for finding the derivative of equations where y is not explicitly given as a function of x.
Strengths of implicit differentiation:
- It allows us to find the derivative when y cannot be expressed explicitly in terms of x.
- It is straightforward and relatively easy to implement.
Weaknesses of implicit differentiation:
- It can be cumbersome and involve more algebraic manipulation compared to finding the derivative using explicit differentiation.
- There is a higher chance of making mistakes due to the additional steps involved.
3. To find the slope of the equation x^2 + xy + y^2 = 3 at the points x = -1, x = 2, and (-1, -1), we can use the same method of implicit differentiation.
Given equation: x^2 + xy + y^2 = 3
Taking the derivative with respect to y:
d/dy (x^2) + d/dy (xy) + d/dy (y^2) = d/dy (3)
0 + x dy/dy + 2y = 0
x + 2y = 0
Here, we have two variables x and y in the equation, and the derivative with respect to y does not isolate dy/dx. Therefore, the method of implicit differentiation fails to find the slope of the equation at the given points.
In this case, we would need to use other techniques to find the slopes, such as explicit differentiation or finding the tangent lines at the given points and calculating their slopes.
To find the slope of the ellipse at the given points when x = ±1 in equation x^2 + 3y^3 = 13, we will differentiate the equation implicitly with respect to x.
1. Find the slope of the ellipse at x = 1:
Differentiating implicitly, we get:
2x + 9y^2 * dy/dx = 0
Solving for dy/dx, we have:
dy/dx = -2x / (9y^2)
Substituting x = 1 into the equation, we have:
dy/dx = -2(1) / (9y^2)
dy/dx = -2 / (9y^2)
2. Find the slope of the ellipse at x = -1:
Differentiating implicitly, we get:
2x + 9y^2 * dy/dx = 0
Solving for dy/dx, we have:
dy/dx = -2x / (9y^2)
Substituting x = -1 into the equation, we have:
dy/dx = -2(-1) / (9y^2)
dy/dx = 2 / (9y^2)
Method:
We used implicit differentiation to find the derivative of y with respect to x. In this method, we treated y as a function of x and differentiated both sides of the equation with respect to x, using the chain rule where necessary.
Strengths of implicit differentiation:
- It allows us to find the derivative of a function defined implicitly.
- It enables us to find the slope of curves defined by equations with both x and y terms.
Weaknesses of implicit differentiation:
- It can be more complex and algebraically challenging compared to explicit differentiation.
- It may not always be possible to solve for dy/dx explicitly in terms of x and y.
3. To find the slopes of the equation x^2 + xy + y^2 = 3 at the given points, we will apply the same method of implicit differentiation and evaluate dy/dx.
Differentiating implicitly, we get:
2x + (x * dy/dx) + 2y * dy/dx + 2y = 0
Simplifying further,
(3x + 2y) * dy/dx = -2x - 2y
dy/dx = (-2x - 2y) / (3x + 2y)
a) For the point x = -1:
Substituting x = -1 into the equation, we have:
dy/dx = (-2(-1) - 2y) / (3(-1) + 2y)
dy/dx = (2 - 2y) / (-3 + 2y)
b) For the point (2, -1):
Substituting x = 2 and y = -1 into the equation, we have:
dy/dx = (-2(2) - 2(-1)) / (3(2) + 2(-1))
dy/dx = (-4 + 2) / (6 - 2)
dy/dx = -2 / 4
dy/dx = -1/2
The method of implicit differentiation can be used to find the slopes of the given equation at the points (-1, -1) and (2, -1).
1. To find the slope of the ellipse at the points x = 1 and x = -1, we need to calculate the derivative of the equation with respect to y. Let's differentiate the given equation:
x^2 + 3y^3 = 13
Differentiating both sides with respect to y:
2x(dx/dy) + 9y^2(dy/dx) = 0
We want to find dy/dx, which represents the slope of the ellipse at the given points. Rearranging the equation, we have:
dy/dx = -(2x(dx/dy))/(9y^2)
Substituting x = 1 and x = -1 into the equation will give us the slope of the ellipse at those points.
2. The method used to find the slope in part 1 is taking the derivative of the equation with respect to y and finding the value of dy/dx. The strength of this method is that it allows us to determine the slope at specific points of the ellipse. However, the weakness of this method is that it only gives us the local slope at those specific points and does not provide the complete picture of how the ellipse behaves.
3. Let's use the same method to find the slope of the equation x^2 + xy + y^2 = 3 at points -1, 2, and (-1, -1). First, we need to differentiate the equation with respect to y:
2x + x(dy/dx) + 2y(dy/dx) = 0
Rearranging the equation, we have:
dy/dx = -(2x + 2y(dy/dx))/(x)
Now, let's substitute the x values to find the slope:
For x = -1:
dy/dx = -(2(-1) + 2y(dy/dx))/(-1) = -(2 - 2y(dy/dx))/(-1)
For x = 2:
dy/dx = -(2(2) + 2y(dy/dx))/(2) = -(4 + 2y(dy/dx))/(2)
For the point (-1, -1), we have x = -1 and y = -1:
dy/dx = -(2(-1) + 2(-1)(dy/dx))/(-1) = -(2 + 2(dy/dx))/(-1)
The method used to find the slope of the ellipse at x = 1 and x = -1 can be applied to the equation x^2 + xy + y^2 = 3 as well. However, when solving for the slope at specific points (-1, 2) and (-1, -1), we need to substitute the given x and y values into the equation.