Describe the slope of the tangent line to the curve defined by x 2 + xy - y 2 = 5 when x = 2.
positive
negative
zero
undefined
2 x dx + x dy + y dx - 2 y dy = 0
dy ( x-2y) = - dx ( 2x + y)
dy/dx = (2 x + y)/(2y-x) = slope
at x = 2
4 + 2 y - y^2 = 5
y^2 - 2 y + 1 = 0
(y -1)^2 = 0
y = 1
so
dy/dx = (5)/0 undefined
To find the slope of the tangent line to the curve, we need to find the derivative of the equation at the point where x = 2.
First, let's rearrange the equation into a more manageable form:
x^2 + xy - y^2 = 5
Now, we can take the derivative of both sides of the equation with respect to x:
d/dx (x^2 + xy - y^2) = d/dx (5)
Using the product rule and chain rule, we get:
2x + y + x(dy/dx) - 2y(dy/dx) = 0
Now, we can substitute the value x = 2 into the equation:
2(2) + y + 2(dy/dx) - 2y(dy/dx) = 0
Simplifying further:
4 + y + 2(dy/dx) - 2y(dy/dx) = 0
Rearranging the terms:
2(dy/dx) - 2y(dy/dx) = y - 4
Factoring out (dy/dx):
(2 - 2y)(dy/dx) = y - 4
Now, we can solve for dy/dx:
dy/dx = (y - 4) / (2 - 2y)
To find the slope of the tangent line at the point where x = 2, we substitute x = 2 into the equation and solve for dy/dx:
dy/dx = (y - 4) / (2 - 2y)
Therefore, the slope of the tangent line to the curve when x = 2 is given by the equation dy/dx = (y - 4) / (2 - 2y).
To determine whether the slope is positive, negative, zero, or undefined, we need to evaluate this equation further. We can plug in different values of y to see what the slope is.
For example, if we plug in y = 0, we get:
dy/dx = (0 - 4) / (2 - 2 * 0) = -4/2 = -2
Since the slope is -2, it is negative.