1. At what point does the normal line to the curve x^2 - XY + Y^2 = 3 at the point (-1,1) intersect the curve again?
2. Find the constants A, B so that if Y=A*sin X + B cos X, then Y satisfies the differential equation Y" + 2Y = 0.
3. Find the points on he graph of Y = e^x - e^2x at which that tangent line is horizontal.
Thank you.
1. To find the point at which the normal line to the curve intersects the curve again, we need to find two things: the equation of the normal line at the point (-1,1) and the intersection point of this line with the curve.
First, we find the derivative dy/dx of the curve x^2 - XY + Y^2 = 3 using implicit differentiation. Taking the derivative of both sides of the equation with respect to x, we get:
2x - (x(dy/dx) + y) + 2y(dy/dx) = 0
Next, we substitute the coordinates (-1, 1) into this equation to find the slope of the tangent line at that point. This will give us the derivative, dy/dx, which represents the slope of the tangent line at that point.
2(-1) - (-1)(dy/dx) + 2(1)(dy/dx) = 0
-2 + dy/dx + 2(dy/dx) = 0
3(dy/dx) - 2 = 0
dy/dx = 2/3
Since the normal line is perpendicular to the tangent line, we know that the slope of the normal line is the negative reciprocal of the slope of the tangent line. So the slope of the normal line is -3/2.
We plug in the point (-1, 1) and the slope of the normal line (-3/2) into the equation y - y1 = m(x - x1), where (x1, y1) is the point of tangency and m is the slope of the normal line:
y - 1 = (-3/2)(x - (-1))
y - 1 = (-3/2)(x + 1)
2y - 2 = -3(x + 1)
2y - 2 = -3x - 3
2y = -3x - 1
y = (-3/2)x - 1/2
Next, we substitute this equation for y into the original curve equation: x^2 - x(-3/2)x - 1/2)^2 = 3
This gives us a quadratic equation in x:
x^2 + (3/2)x^2 + 3x + (1/2)^2 - 3 = 0
(5/2)x^2 + 3x + 1/4 - 3 = 0
(5/2)x^2 + 3x - 11/4 = 0
Now, we can solve this quadratic equation to find the x-coordinates of the points where the normal line intersects the curve again.
2. To find the constants A and B so that Y = A*sin X + B*cos X satisfies the differential equation Y" + 2Y = 0, we need to find the second derivative of Y.
First, we find the first derivative of Y with respect to X:
dY/dX = A*cos X - B*sin X
Next, we find the second derivative of Y with respect to X:
d^2Y/dX^2 = -A*sin X - B*cos X
Now, we substitute these derivatives into the differential equation Y" + 2Y = 0 and simplify:
(-A*sin X - B*cos X) + 2(A*sin X + B*cos X) = 0
-A*sin X - B*cos X + 2A*sin X + 2B*cos X = 0
A*sin X + B*cos X = 0
Since this equation should hold for all values of X, we can conclude that both the coefficient of sine (A) and the coefficient of cosine (B) must be equal to zero. Therefore, A = 0 and B = 0.
3. To find the points on the graph of Y = e^x - e^(2x) at which the tangent line is horizontal, we need to find the derivative of Y with respect to X and then find the X-values at which the derivative is equal to zero.
First, we find the derivative dY/dX:
dY/dX = e^x - 2e^(2x)
Next, we set the derivative equal to zero and solve for X:
e^x - 2e^(2x) = 0
To solve this equation, we can factor out e^x:
e^x(1 - 2e^x) = 0
This equation is satisfied when either e^x = 0 (which is not possible) or 1 - 2e^x = 0. Solving the second equation for e^x, we get:
1 - 2e^x = 0
2e^x = 1
e^x = 1/2
x = ln(1/2)
So the tangent line is horizontal at the point (ln(1/2), e^(ln(1/2)) - e^(2ln(1/2))) on the graph of Y = e^x - e^(2x).