To determine the slope of the graph at the point (3,1), we need to find the derivative of the equation with respect to x and y, and then substitute the values x=3 and y=1 into the derivatives.
First, let's find the derivative with respect to x. We can rewrite the equation as:
3(x^2+y^2)^2 - 100xy = 0
Differentiating both sides with respect to x:
d/dx [3(x^2+y^2)^2] - d/dx [100xy] = 0
Applying the chain rule, we get:
6(x^2+y^2)(2x) - 100y - 100xdx/dx = 0
The term dx/dx simplifies to 1. Rearranging the equation, we have:
12x(x^2+y^2) - 100y = 0 ...(1)
Now, let's find the derivative with respect to y. We differentiate both sides of the original equation:
d/dy [3(x^2+y^2)^2] - d/dy [100xy] = 0
Again applying the chain rule, we have:
6(x^2+y^2)(2y) - 100x - 100ydy/dy = 0
The term dy/dy simplifies to 1. Rearranging the equation, we get:
12y(x^2+y^2) - 100x = 0 ...(2)
Now, we substitute x=3 and y=1 into equations (1) and (2) to find the slope at the point (3,1).
From equation (1): 12(3)(3^2+1^2) - 100(1) = 0
108(10) - 100 = 0
1080 - 100 = 0
980 ≠ 0
From equation (2): 12(1)(3^2+1^2) - 100(3) = 0
12(1)(10) - 300 = 0
120 - 300 = 0
-180 ≠ 0
Since neither equation (1) nor equation (2) simplifies to zero when x=3 and y=1, there is no unique slope at the point (3,1) for the given equation.
Determine the slope of the graph 3(x^2+y^2)^2=100xy at the point (3,1)
5 answers
That's just a very horrible attempt by the bot
3(x^2+y^2)^2=100xy
6(x^2 + y^2)(2x + 2y dy/dx) = 100y + 100x dy/dx
sub in the point (3,1)
6(9+1)(6 + 2y dy/dx) = 100 + 300 dy/dx
60(6 + 2dy/dx) = 100 + 300 dy/dx
36 + 12dy/dx = 10 + 30 dy/dx
18 dy/dx = 26
dy/dx = 26/18 = 13/9
3(x^2+y^2)^2=100xy
6(x^2 + y^2)(2x + 2y dy/dx) = 100y + 100x dy/dx
sub in the point (3,1)
6(9+1)(6 + 2y dy/dx) = 100 + 300 dy/dx
60(6 + 2dy/dx) = 100 + 300 dy/dx
36 + 12dy/dx = 10 + 30 dy/dx
18 dy/dx = 26
dy/dx = 26/18 = 13/9
Apologies for the previous incorrect response. Let's correct it:
To determine the slope of the graph at the point (3,1), we can find the derivative of the equation with respect to x and y, then substitute the values x=3 and y=1 into the derivatives.
We start by differentiating both sides of the equation with respect to x:
6(x^2+y^2)(2x) = 100y + 100x (dy/dx)
Next, we differentiate each term with respect to y, treating x as a constant:
12x(x^2+y^2) - 100 = 100x (dy/dx)
Since we want to find the slope at the point (3,1), we substitute x=3 and y=1 into the derivative equation:
12(3)(3^2+1^2) - 100 = 100(3) (dy/dx)
108(10) - 100 = 300 (dy/dx)
dy/dx = (1080 - 100) / 300
dy/dx = 980 / 300
Simplifying, we get:
dy/dx = 49 / 15
Therefore, the slope of the graph at the point (3,1) is 49/15.
To determine the slope of the graph at the point (3,1), we can find the derivative of the equation with respect to x and y, then substitute the values x=3 and y=1 into the derivatives.
We start by differentiating both sides of the equation with respect to x:
6(x^2+y^2)(2x) = 100y + 100x (dy/dx)
Next, we differentiate each term with respect to y, treating x as a constant:
12x(x^2+y^2) - 100 = 100x (dy/dx)
Since we want to find the slope at the point (3,1), we substitute x=3 and y=1 into the derivative equation:
12(3)(3^2+1^2) - 100 = 100(3) (dy/dx)
108(10) - 100 = 300 (dy/dx)
dy/dx = (1080 - 100) / 300
dy/dx = 980 / 300
Simplifying, we get:
dy/dx = 49 / 15
Therefore, the slope of the graph at the point (3,1) is 49/15.
geeesshhhh, I gave you the solution !!!!
Apologies for the confusion. You are correct. From your solution, we have:
dy/dx = 13/9
Therefore, the slope of the graph at the point (3,1) is indeed 13/9. Thank you for providing the correct solution.
dy/dx = 13/9
Therefore, the slope of the graph at the point (3,1) is indeed 13/9. Thank you for providing the correct solution.