find all points on the curve x^2y^2+xy=2 where the slope of the tangent line is -1
I know that you have to find the derivative when it equals -1
i got
y'=(2x*y^2-y)/(2x^2*y+x)
(i don't know if it is correct or not)
but i don't know how to go from here. any help would be good
Thanks
try changing variables. let z=xy in the origianal equation, solve for z
z^2+z-2=0
(z+2)(z-1)=0
z=1 or -2 or z is a constant.
Then z=xy or
z'=y+xy'=0
but y'=1
y=-x and xy=-1 or 1
xy=-x^2
-2=-x^2
x=sqrt2, y=-x =-sqrt2
check: original line 4-2=2 checks.
1=-x^2
x=i, y=-x=-i
check: originalline 1=1=2 checks.
solutions:
pointA: x=i,y=-1
PointB: x=sqrt2, y=-sqrt2
Can you explain using implicit differention?
No, I tried it that way, it bogged down in a third degree equation after I substituted, so I gave up that way.
To find all points on the curve where the slope of the tangent line is -1, you have correctly started by finding the derivative of the curve. However, there seems to be a mistake in your derivative calculation.
Let's proceed step by step to find the derivative correctly:
1. Start with the equation of the curve: x^2y^2 + xy = 2.
2. Differentiate both sides of the equation with respect to x using implicit differentiation.
- For the left side, apply the Product Rule: d/dx (x^2y^2) = 2xy^2 + 2x^2yy'.
- For the right side, the derivative of a constant (2) is zero.
- The derivative of xy is y + xy'.
Now, collect the terms with y' on one side to get the derivative:
2xy^2 + 2x^2yy' + y + xy' = 0
Next, let's rearrange the equation to solve for y':
2x^2yy' + xy' = -2xy^2 - y
Factor out y' from the left side and y from the right side:
y'(2x^2y + x) = -y(2xy + 1)
Divide both sides by (2x^2y + x) and simplify:
y' = -y(2xy + 1)/(2x^2y + x)
Now, you want to find the points on the curve where the slope of the tangent line is -1. This means that the derivative, y', must be equal to -1. So, we can set:
-y(2xy + 1)/(2x^2y + x) = -1
To solve this equation, we can multiply both sides by (2x^2y + x) to eliminate the denominator:
-y(2xy + 1) = - (2x^2y + x)
Expand the left side:
-2x^2y^2 - y = -2x^2y - x
Move all terms to one side:
-2x^2y^2 - y + 2x^2y + x = 0
Combine like terms:
-2x^2y^2 + 2x^2y - y + x = 0
Now, we have a quadratic equation in terms of y. To solve this, we can either factor or use the quadratic formula.
Since we're looking for all points on the curve, we can try factoring this expression:
2x^2y^2 - 2x^2y + x - y = 0
y(2x^2y - 2x^2) - (x - y) = 0
y(2x^2(y - 1)) - (x - y) = 0
Factoring out the common factor:
(y - 1)(2x^2y - x + 1) = 0
Now, we have two equations:
1) y - 1 = 0
2) 2x^2y - x + 1 = 0
Solving equation 1) gives y = 1.
To solve equation 2), you can either use the quadratic formula or rearrange it to find the specific values of x and y. This will give you the points on the curve where the slope of the tangent line is -1.