The circle defined by the equation x^2 + y^2 = 18 has two points where the slope of its tangent line is m=1. Find those points.
Good idea. However, in an attempt to use your math, and also to apply to a possibly more general problem later, try implicit differentiation:
x^2 + y^2 = 18
2x + 2yy' = 0
y' = -x/y
So, if -x/y = 1 and x^2+y^2 = 18,
2x^2 = 18
x = ±3
Now, figuring y should not be too hard.
Well, well, well, looks like this circle wants to flirt with some tangents! Who can blame it? Okay, let me get my clown calculator out and see if I can help you find those saucy tangent points.
To find the points where the slope of the tangent line is m=1, we need to find the derivative of the circle equation. Taking the derivative of x^2 + y^2 = 18, we get 2x + 2yy' = 0.
Since we want the slope of the tangent line to be m = 1, we can set y' (the derivative of y with respect to x) equal to 1 and solve for x. So, 2x + 2y(1) = 0 simplifies to:
2x + 2y = 0.
Now, we can substitute this back into the original circle equation:
x^2 + y^2 = 18.
Substituting 2x + 2y = 0, we get:
x^2 + (-x)^2 = 18.
Simplifying:
2x^2 = 18.
Dividing both sides by 2 gives:
x^2 = 9.
Taking the square root of both sides, we find two possible values for x: x = 3 and x = -3.
Now, let's substitute these x-values back into 2x + 2y = 0 to find the corresponding y-values.
For x = 3: 2(3) + 2y = 0.
Simplifying:
6 + 2y = 0,
2y = -6,
y = -3.
So, one of the tangent points is (3, -3).
For x = -3: 2(-3) + 2y = 0.
Simplifying:
-6 + 2y = 0,
2y = 6,
y = 3.
Therefore, the other tangent point is (-3, 3).
And there you have it! Clown Bot has successfully located the two points where the slope of the tangent line is m=1 for the circle x^2 + y^2 = 18. Keep up the mathematical mischief!
To find the points on the circle where the slope of the tangent line is m=1, we need to find the points where the derivative of the equation of the circle is equal to 1.
Given the equation of the circle x^2 + y^2 = 18, we can differentiate both sides of the equation with respect to x:
2x + 2y(dy/dx) = 0
To find the points where the slope of the tangent line is 1, we set dy/dx = 1 and solve for x.
2x + 2y(1) = 0
2x + 2y = 0
y = -x
Substituting y = -x back into the equation of the circle, we get:
x^2 + (-x)^2 = 18
x^2 + x^2 = 18
2x^2 = 18
x^2 = 9
x = ±sqrt(9)
x = ±3
So the x-coordinates of the points are x = 3 and x = -3. Substituting these values into y = -x, we get:
For x = 3, y = -3
For x = -3, y = 3
Therefore, the two points where the slope of the tangent line to the circle is m=1 are (3, -3) and (-3, 3).
To find the points where the slope of the tangent line is equal to 1 on the circle x^2 + y^2 = 18, we need to use calculus.
Step 1: Differentiate the equation of the circle with respect to x to find the derivative of the circle's equation:
d/dx (x^2 + y^2) = d/dx(18)
2x + 2y(dy/dx) = 0
Step 2: Rearrange the equation to solve for dy/dx (the slope):
dy/dx = -x/y
Step 3: Substitute dy/dx = 1 into the equation and solve for x and y simultaneously:
-1 = -x/y
Multiplying both sides by y,
y = -x
Step 4: Substitute y = -x into the equation of the circle to find the corresponding x-coordinate:
x^2 + (-x)^2 = 18
2x^2 = 18
x^2 = 9
x = ±3
Step 5: Substitute the values of x into the equation y = -x to find the corresponding y-coordinates:
For x = 3, y = -3
For x = -3, y = 3
Therefore, the two points on the circle where the slope of the tangent line is 1 are (3, -3) and (-3, 3).