Can someone show how this question is solved.
Consider the curve given by the equation
2y^3 + y^2 - y^5 = x^4 -2x^3 +x^2
Find all points at which the tangent line to the curve is horizontal or vertical.
Thanks!
use implicit differentiation, ....
2y^3 + y^2 - y^5 = x^4 -2x^3 +x^2
6y^2 (dy/dx) + 2y (dy/dx) - 5y^4 (dy/dx) = 4x^3 - 6x^2 + 2x
dy/dx = (4x^3 - 6x^2 + 2x)/(6y^2 + 2y - 5y^4)
for a horizontal tangent, dy/dx = 0 ---> the numerator is zero
5x^3 - 6x^2 + 2x = 0
x(5x^2 - 6x + 2) = 0
the quadratic has no real solutions, so x = 0
then if x = 0
2y^3 + y^2 - y^5 = 0-0+0
y^2(2y + 1 - y^3) = 0
y = 0 , or y^3 - 2y - 1 = 0 ---> one point is (0,0)
y^3 - 2y - 1 = 0, easy to see that y = -1
using synthetic division by y+1 , I was left with
y^2 - y - 1 = 0 which solves to get
y = (1 ± √5)/2 ,..... mmmh, I see the golden ratio in there
So you got 3 points where the tangent is horizontal.
for vertical tangents, the denominator is zero
6y^2 + 2y - 5y^4 = 0
y(6y + 2 - 5y^3) = 0
y = 0 or 5y^3 - 6y - 2 = 0
so there is the (0,0) again, but 5y^3 - 6y - 2 = 0 is a mess to solve (unless I made an arithmetic error)
So I went with Wolfram:
http://www.wolframalpha.com/input/?i=5y%5E3+-+6y+-+2+%3D+0
to get 3 more points where the tangent is vertical
Might as well look at the original graph to see if this makes sense.
http://www.wolframalpha.com/input/?i=2y%5E3+%2B+y%5E2+-+y%5E5+%3D+x%5E4+-2x%5E3+%2Bx%5E2
Wow!
To find all points at which the tangent line to the curve is horizontal or vertical, we need to find the values of x and y that satisfy the given equation and also satisfy the condition of a horizontal or vertical tangent line.
Step 1: Differentiate the equation with respect to x to find the derivative of y with respect to x (dy/dx).
First, let's rearrange the equation so that it is in the form y = f(x):
2y^3 + y^2 - y^5 = x^4 - 2x^3 + x^2
Rearranging gives:
y^5 + 2y^3 + y^2 - x^4 + 2x^3 - x^2 = 0
Differentiating both sides of the equation with respect to x:
5y^4(dy/dx) + 6y^2(dy/dx) + 2y(dy/dx) - 4x^3 + 6x^2 - 2x = 0
Step 2: Determine the conditions for horizontal and vertical tangent lines.
A horizontal tangent line occurs when dy/dx = 0, and a vertical tangent line occurs when the derivative is undefined, which happens when dy/dx is infinite or does not exist.
Step 3: Set dy/dx equal to zero and solve for y.
Setting the derivative equal to zero and solving for y:
5y^4(dy/dx) + 6y^2(dy/dx) + 2y(dy/dx) - 4x^3 + 6x^2 - 2x = 0
(5y^4 + 6y^2 + 2y)(dy/dx) = 4x^3 - 6x^2 + 2x
Since we're looking for the points at which the tangent line is horizontal or vertical, dy/dx should be zero:
5y^4 + 6y^2 + 2y = 0
Step 4: Solve the equation 5y^4 + 6y^2 + 2y = 0 for y.
To solve this equation, we factor out a y:
y(5y^3 + 6y + 2) = 0
Setting both factors equal to zero:
y = 0 or 5y^3 + 6y + 2 = 0
Step 5: Solve the cubic equation 5y^3 + 6y + 2 = 0 for y.
To solve the cubic equation, you can use numerical or graphical methods, or apply techniques such as the rational root theorem, synthetic division, or Newton's method.
Solving this equation can be intricate depending on the coefficients. If a numerical solution is desired, it is recommended to use a graphing calculator, an online solver, or specialized software.
Once you have the solutions for y, substitute them back into the original equation to find the corresponding values of x. These points will be the ones where the tangent line is horizontal or vertical on the curve.