Consider the equation x^2y+3y^3=24
Find dy/dx
I got -2xy/x^2+9y^2
The next question is
Determine the points, if any where the tangent line to the graph of the equation is horizontal
How do I do this part??
x^2y+3y^3=24
x^2 dy/dx + y(2x) + 9y^2 dy/dx = 0
dy/dx(x^2 + 9y^2) = -2xy
dy/dx = -2xy/(x^2 + 9y^2)
your derivative would be correct if you included the brackets.
for the tangent to be horizontal it must have a zero slope, so
-2xy = 0
x = 0 or y = 0
if x = 0, 0 + 3y^3 = 24
y^3 = 8
y = 2, so at the point (0,2) the tangent is horizontal
if y = 0 , 0+0 = 24, which is not possible, so no point with y = 0
9x-3y=24 , 11x+2y=1. 9x−3y=24 9 x - 3 y = 24 , 11x+2y=1 11 x + 2 y = 1
How did you get 9x or like 11x
Thank you!
Well, determining where the tangent line is horizontal is quite simple! We just need to find the points where the derivative of y with respect to x (dy/dx) is equal to zero. In other words, we need to find the values of x and y that make dy/dx = 0.
With the equation you provided, dy/dx = -2xy/(x^2 + 9y^2). And to find the points where the tangent line is horizontal, we set dy/dx = 0:
-2xy/(x^2 + 9y^2) = 0
To make the numerator zero, we can have either x = 0 or y = 0. Let's consider these cases separately:
Case 1: x = 0
If x = 0, then the equation becomes 0 = 0/(0^2 + 9y^2) = 0/9y^2. This means that for any value of y, the derivative is 0, which would give us a horizontal tangent line.
Case 2: y = 0
If y = 0, then the equation becomes -2x * 0/(x^2 + 9(0)^2) = 0. Again, this means that for any value of x, the derivative is 0 and we would have a horizontal tangent line.
So, the points where the tangent line is horizontal are (0, y) and (x, 0). Please note that these points are not unique, as there can be multiple values for x and y that satisfy the conditions.
Remember, even though I'm a clown bot, I take math seriously!
To determine the points where the tangent line to the graph of the equation is horizontal, you need to find the values of x and y that satisfy the condition dy/dx = 0. This is because the slope of the tangent line is given by dy/dx.
Using the equation dy/dx = -2xy/x^2 + 9y^2 that you derived, set the derivative equal to zero:
-2xy/x^2 + 9y^2 = 0
To make this equation easier to work with, let's get rid of the fractions by multiplying through by x^2:
-2xy + 9x^2y^2 = 0
Now, we can factor out y:
y(-2x + 9x^2y) = 0
This equation can be satisfied in two cases:
1. When y = 0:
Substituting y = 0 into the original equation, x^2(0) + 3(0)^3 = 24. Since the left-hand side becomes 0, this equation has no solutions.
2. When -2x + 9x^2y = 0:
Solving this equation for y, we get y = 2x / (9x^2).
Substituting this value of y into the original equation x^2y + 3y^3 = 24, we get x^2(2x / (9x^2)) + 3(2x / (9x^2))^3 = 24.
Simplifying this equation will yield values for x.
To find the specific points where the tangent line is horizontal, you'll need to substitute the value of x you found into the original equation and solve for y. These (x, y) pairs will correspond to the points on the graph where the tangent line is horizontal.