a)
If x3 + y3 - xy2 = 5, find dy/dx.
(b)
Using your answer to part (a), make a table of approximate y-values of points on the curve near x = 1, y = 2. Include x = 0.96, 0.98, 1, 1.02, 1.04.
(c)
Find the y-value for x = 0.96 by substituting x = 0.96 in the original equation and solving for y using a computer or calculator. Compare with your answer in part (b).
(d)
Find all points where the tangent line is horizontal or vertical.
a) Assuming you meant
x^3 + y^3 - x y^2 = 5
3x^2 + 3y^2 dy/dx - x(2y)dy/dx - y^2 = 0
dy/dx(3y^2 - 2xy) = y^2 - 3x^2
dy/dx = (y^2 - 3x^2)/(3y^2 - 2xy)
b) just arithmetic, you do it
c) yours
d) for a horizontal tangent, dy/dx = 0
then y^2 - 3x^2 = 0
y^2 = 3x^2
y = ± x√3
sub into the first:
x^3 + 3√3x^3 - x(x√3) = 5
x^3 + 3√3 x^3 - √3 x^2 - 5 = 0
Nastiness!!!!
wolfram says x = 1.03412 correct to 5 decimals
now sub that into the original and find y, equally nasty.
for vertical asymptote , denominator of dy/dx = 0
3y^2 - 2xy) = 0
y(3y - 2x) = 0
y = 0 , that is not bad
or
y = 2x/3
more mess ahead
I suggest you use the Wolfram page or your built-in equation solver of your fancy calculator to solve these equations.
(it might be a good idea to check my work, lately I have been making silly errors by not writing out the solution first)
a) Well, it looks like we have a math problem here. Let's find dy/dx, also known as the derivative. Just give me a second to dust off my mathematician hat... Ahem. *puts on a hat made of numbers and symbols* Alright, let's do this!
To find dy/dx, we need to differentiate the equation x^3 + y^3 - xy^2 = 5 with respect to x. Are you ready? Here we go!
b) Now that we have found the derivative, let's use it to make a table of approximate y-values near x = 1, y = 2. Since we don't want any round numbers or whole numbers to give us accurate results, let's get a little fancy with some decimal numbers. How about x = 0.96, 0.98, 1, 1.02, and 1.04? Let's crunch those numbers, shall we?
c) So, we already got approximate y-values for x = 0.96 in part (b), but let's see how accurate we were. To find the exact y-value for x = 0.96, we can substitute it into the original equation and solve for y using a computer or calculator. Let's compare it to our earlier approximation, shall we?
d) Now, let's find all the points where the tangent line is horizontal or vertical. In other words, let's find those magical points where the slope is either zero or undefined. Are you ready for this exciting journey into the land of straight lines?
Alright, I hope my attempts at humor haven't scared you away. Now, let's tackle those math problems together!
(a) To find dy/dx, we can implicitly differentiate the equation x^3 + y^3 - xy^2 = 5 with respect to x.
Differentiating both sides of the equation using the chain rule, we get:
3x^2 + 3y^2 * dy/dx - y^2 - 2xy * dy/dx = 0
Rearranging the equation to solve for dy/dx, we have:
dy/dx = (y^2 - 3x^2) / (3y^2 - 2xy)
(b) To make a table of approximate y-values of points on the curve near x = 1, y = 2, we will use the expression for dy/dx we found in part (a) and approximate values of x.
- For x = 0.96:
Substitute x = 0.96 into the equation x^3 + y^3 - xy^2 = 5 and solve for y.
Calculate dy/dx using the expression found in part (a).
Determine the approximate y-value.
- For x = 0.98, 1, 1.02, and 1.04:
Calculate dy/dx using the expression found in part (a).
Use the initial point (x = 1, y = 2) and the calculated dy/dx to estimate the y-values.
(c) Substituting x = 0.96 in the original equation x^3 + y^3 - xy^2 = 5, we have:
(0.96)^3 + y^3 - (0.96)y^2 = 5
Solving this equation for y using a computer or calculator, we can find the y-value for x = 0.96.
Comparing this value with the approximate value from part (b) will allow us to check the accuracy of the estimation.
(d) To find the points where the tangent line is horizontal or vertical, we need to identify the values of x and y for which dy/dx is zero or undefined.
- Horizontal tangent: dy/dx = 0
Set (y^2 - 3x^2) = 0 and solve for x and y.
- Vertical tangent: dy/dx is undefined
Set (3y^2 - 2xy) = 0 and solve for x and y.
The solutions to these equations will give the points where the tangent line is either horizontal or vertical.
(a) To find dy/dx, we need to differentiate the given equation with respect to x. Differentiating x3 gives us 3x^2, differentiating y3 gives us 3y^2 * dy/dx, differentiating -xy^2 gives us -y^2 - 2xy * dy/dx. Setting this derivative equal to zero, we have:
3x^2 - y^2 - 2xy * dy/dx = 0.
Rearranging this equation, we get:
dy/dx = (3x^2 - y^2) / (2xy).
(b) Using the expression we found in part (a), we can compute the approximate y-values for the given x-values. Let's substitute each x-value into the expression for dy/dx and solve for y.
For x = 0.96:
dy/dx = (3(0.96)^2 - y^2) / (2(0.96)y).
We need to solve this equation for y using a computer or calculator.
For x = 0.98:
dy/dx = (3(0.98)^2 - y^2) / (2(0.98)y).
For x = 1:
dy/dx = (3(1)^2 - y^2) / (2(1)y).
For x = 1.02:
dy/dx = (3(1.02)^2 - y^2) / (2(1.02)y).
For x = 1.04:
dy/dx = (3(1.04)^2 - y^2) / (2(1.04)y).
(c) To find the y-value for x = 0.96, we substitute x = 0.96 into the original equation and solve for y:
(0.96)^3 + y^3 - 0.96y^2 = 5.
We need to solve this equation using a computer or calculator.
(d) To find the points where the tangent line is horizontal or vertical, we need to find the values of x and y that make dy/dx equal to zero or undefined.
For a horizontal tangent line:
Set dy/dx = 0 and solve for x and y.
For a vertical tangent line:
Set the denominator of the expression for dy/dx equal to zero and solve for x and y.
By finding these points, we can determine where the tangent line is either horizontal or vertical.