Given x^2+y^2=4, find the equation of the tangent line at the point (-1,sq.rt.3). Then, at what point is the slope 0? What point is the slope -1?
I have no clue what to do!
Differentiate the equation: 2x+2ydy/dx=0 2ydy/dx=-2x dy/dx=-x/y dy/dx is the gradient m substitute for x and y m=1/sqrt3 equation for tangent line is y-y1=m(x-x1) where x1 and y1 are -1 and sqrt3 substitute it in equation b)when dy/dx=0 x=0 c)when dy/dx equal -1 x=y
and also draw a picture of the circle of radius 2 and center at (0,0) so you get an idea what is going on.
To find the equation of the tangent line at a given point on a curve, you need to do the following steps:
1. Find the derivative of the function representing the curve.
The given equation, x^2 + y^2 = 4, represents a circle with radius 2 centered at the origin. We can rearrange it to y^2 = 4 - x^2, and then take the square root of both sides to get y = sqrt(4 - x^2).
2. Differentiate the function with respect to x to find the derivative.
Using the power rule for differentiation, the derivative of sqrt(4 - x^2) with respect to x can be found. It is dy/dx = (-x) / sqrt(4 - x^2).
3. Plug in the x-coordinate of the given point (-1, sqrt(3)) into the derivative to find the slope of the tangent line at that point.
Substituting x = -1 into the derivative, we get dy/dx = (-(-1)) / sqrt(4 - (-1)^2) = 1 / sqrt(3).
4. Use the point-slope form of a linear equation to write the equation of the tangent line.
Using the point-slope form with the given point (-1, sqrt(3)) and the slope 1/sqrt(3), we get y - sqrt(3) = (1/sqrt(3))(x + 1). Simplifying, we get y = (1/sqrt(3))x + (sqrt(3) + 1/sqrt(3)).
Now, let's find the points where the slope is 0 and -1.
5. To find the point where the slope is 0, we set dy/dx = 0 and solve for x.
0 = (-x) / sqrt(4 - x^2)
Multiplying both sides by sqrt(4 - x^2), we get 0 = -x.
Hence, x = 0. Plugging x = 0 into the original equation (x^2 + y^2 = 4), we find y = 2 and y = -2.
Therefore, the point where the slope is 0 is (0, 2) and (0, -2).
6. To find the point where the slope is -1, we set dy/dx = -1 and solve for x.
-1 = (-x) / sqrt(4 - x^2)
Multiplying both sides by sqrt(4 - x^2), we get -sqrt(4 - x^2) = -x.
Squaring both sides, we get 4 - x^2 = x^2. Rearranging, we have 2x^2 = 4, and x = ±√2. Plugging these values into the original equation, we find y = ±√(4 - 2) = ±√2.
Therefore, the points where the slope is -1 are (√2, √2), (√2, -√2), (-√2, √2), and (-√2, -√2).