Consider the curve given by the equation y^3+3x^2y+13=0
a.find dy/dx
b. Write an equation for the line tangent to the curve at the point (2,-1)
c. Find the minimum y-coordinate of any point on the curve.
the work for these would be appreciated i don't need the answers.
differentiate implicitly.
3y^2 dy/dx + 3x^2 dy/dx + 6xy + 0 = 0
dy/dx(3y^2 + 3x^2) = -6xy
dy/dx/ = -6xy/(3y^2 + 3x^2)
b) sub in x = 2, to get the slope, then use the grade 9 way to find the equation of a line given the slope and a point.
c) set dy/dx = 0
so -6xy = 0
x = 0 or y = 0
when x = 0, y^3 +1 = -13
y = (-13)^(1/3) = appr. - 2.35
when y = 0, 0 + 0 + 13 = 0 , not possible.
so we have one turning point at (0,-2.35)
what about when x becomes large either + or -
investigate for x = 100
y^3 + 30000y = -13
y = -13/30000 which approaches zero.
the same thing is true if x = -100
so the minimum value of y is -2.35 when x = 0
Well, here's a helpful math joke for you before we dive into the work:
Why was the math book sad?
Because it had too many problems! 😄
Now, onto the problem at hand:
a. To find dy/dx, we'll need to use implicit differentiation. Let's differentiate both sides of the equation with respect to x:
d/dx(y^3 + 3x^2y + 13) = d/dx(0)
Using the chain rule and product rule, we get:
3y^2(dy/dx) + 6xy + 3x^2(dy/dx) = 0
Now, we can isolate dy/dx by moving all the terms with dy/dx to one side:
(3y^2 + 3x^2) (dy/dx) = -6xy
Finally, divide both sides by (3y^2 + 3x^2) to obtain the expression for dy/dx.
b. To find the equation of the tangent line at the point (2,-1), we can use the point-slope form of a line: y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope of the line.
We already have the slope (dy/dx). Now let's plug in the point (2,-1):
y - (-1) = (dy/dx) (x - 2)
Simplifying further, we get the equation of the tangent line.
c. To find the minimum y-coordinate of any point on the curve, we'll need to examine the nature of the curve. More specifically, we need to find the critical points where dy/dx = 0 or is undefined. Then, we can evaluate these points to determine the minimum y-coordinate.
To be honest, finding the exact values of these critical points can be quite tricky, involving solving cubic equations. However, you can use numerical methods (like graphing or approximation techniques) to estimate the value.
Remember, math may not always be fun, but with a dash of humor, it becomes much more enjoyable. Is there anything else I can help you with?
a. To find dy/dx, we differentiate the given equation implicitly with respect to x.
Differentiating y^3 + 3x^2y + 13 = 0 with respect to x, we get:
3y^2(dy/dx) + 6xy + 3x^2(dy/dx) = 0
Now, we can solve for dy/dx:
3y^2(dy/dx) + 3x^2(dy/dx) = -6xy
(dy/dx)(3y^2 + 3x^2) = -6xy
(dy/dx) = -6xy / (3y^2 + 3x^2)
(dy/dx) = -2xy / (y^2 + x^2)
b. To find the equation of the tangent line at the point (2, -1), we need both the slope and a point on the line.
Let's find the slope first. Substituting x = 2 and y = -1 into the expression for dy/dx, we get:
dy/dx = -2(2)(-1) / ((-1)^2 + 2^2)
dy/dx = 4 / 5
Now we can substitute the coordinates of the point (2, -1) and the slope into the point-slope form of a linear equation (y - y1 = m(x - x1)) to find the equation of the tangent line.
c. To find the minimum y-coordinate of any point on the curve, we can set dy/dx equal to zero and solve for y.
Setting dy/dx = 0, we have:
-2xy / (y^2 + x^2) = 0
-2xy = 0
This equation has two possibilities:
1. x = 0: In this case, any value of y would satisfy the equation. Therefore, there is no minimum y-coordinate.
2. y = 0: In this case, the equation becomes 13 = 0, which is not possible. So, y cannot be zero.
Hence, there is no minimum y-coordinate on the curve.
a. To find dy/dx, we need to differentiate the given equation with respect to x. Let's differentiate term by term:
The first term, y^3, can be written as y * y^2. Applying the product rule, we get:
d/dx (y^3) = y * d/dx (y^2) + y^2 * d/dx (y)
The second term, 3x^2y, can be rewritten as 3xy^1 * x. Applying the product rule, we get:
d/dx (3x^2y) = 3xy^1 * d/dx (x) + x * d/dx (3xy^1)
The constant term, 13, differentiates to zero since it does not contain any variables.
Now, let's differentiate each term individually:
d/dx (y^3) = y * d/dx (y^2) + y^2 * d/dx (y)
= y * 2y dy/dx + y^2 dy/dx
= 2y^2 dy/dx + y^2 dy/dx
d/dx (3x^2y) = 3xy^1 * d/dx(x) + x * d/dx(3xy^1)
= 3xy^1 * 1 + x * (3y^1 * d/dx (x) + x * d/dx (y^1))
= 3xy + 3x^2 dy/dx + xy^1 * dy/dx
= 3xy + 3x^2 dy/dx + xy * dy/dx
Now, rearrange the terms and simplify:
2y^2 dy/dx + y^2 dy/dx + 3xy + 3x^2 dy/dx + xy dy/dx = 0
Combine like terms:
(2y^2 + y^2 + 3xy + xy) dy/dx + (3x^2) dy/dx = 0
Factor out dy/dx:
(3x^2 + 3xy + 3y^2) dy/dx = -3y^2 - 4xy
Finally, solve for dy/dx:
dy/dx = (-3y^2 - 4xy) / (3x^2 + 3xy + 3y^2)
b. To find the equation of the tangent line at a point (2, -1), we need the slope of the tangent line and a point on the line.
From part a, we found that dy/dx = (-3y^2 - 4xy) / (3x^2 + 3xy + 3y^2). Substituting x = 2 and y = -1 into this equation will give us the slope of the tangent line at (2, -1).
So, let's substitute these values into dy/dx:
dy/dx = (-3(-1)^2 - 4(2)(-1)) / (3(2)^2 + 3(2)(-1) + 3(-1)^2)
= (-3 + 8) / (12 - 6 - 3)
= 5 / 3
Now we have the slope of the tangent line, which is 5/3.
To find a point on the line, we use the point (2, -1).
Now we can use the point-slope form of the equation of a line to write the equation. The equation will be in the form:
y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope.
Substituting the values, we get:
y - (-1) = (5/3)(x - 2)
y + 1 = (5/3)(x - 2)
This is the equation for the line tangent to the curve at the point (2, -1).
c. To find the minimum y-coordinate, we need to minimize the function y = f(x) obtained from the given equation y^3 + 3x^2y + 13 = 0.
To find the minimum value, we need to differentiate the function with respect to x and set it equal to zero.
Using the quotient rule, we differentiate the function:
dy/dx = [3x^2(d/dx(y)) + y(d/dx(3x^2)) + (y^3 + 3x^2y + 13)(d/dx(y))] / (y^3 + 3x^2y + 13)^2
Setting dy/dx equal to zero, we get:
0 = [3x^2(dy/dx) + y(6x) + (y^3 + 3x^2y + 13)(dy/dx)] / (y^3 + 3x^2y + 13)^2
Simplifying the equation and isolating dy/dx, we get:
dy/dx = -y(6x) / (3x^2(dy/dx) + 3x^2y + y^3 + 13)
Now, substitute the coordinates of the points into the obtained derivative equation and solve for dy/dx to find the minimum y-coordinate.