(a) Find dy/dx given that x^2+y^2-4x+7y=15.
(b) Under what conditions on x and/or y is the tangent line to this curve horizontal? Vertical?
For part A, I simplifid it to
-(2x+4/2y+7)
For part B,I am confused on what to do?
Assuming the y' formula is correct, the tangent is zero when y' = 0 (x = 2) and vertical when the denominator of y' is zero (y = -7/2). The corresponding y and x values can be solve for using the original equation.
2x+2y.y'-4+7y'=0
y'(2y+7)=(4-2x)
y'=(4-2x)/(2y+7) as you found:)
For part A, the first step is to rearrange the given equation into the standard form of a circle:
x^2 + y^2 - 4x + 7y = 15
Rearranging terms, we have:
(x^2 - 4x) + (y^2 + 7y) = 15
Completing the square for x terms by adding (4/2)^2 = 4 to both sides:
(x^2 - 4x + 4) + (y^2 + 7y) = 15 + 4
(x - 2)^2 + (y^2 + 7y) = 19
Completing the square for y terms by adding (7/2)^2 = 49/4 to both sides:
(x - 2)^2 + (y^2 + 7y + 49/4) = 19 + 49/4
(x - 2)^2 + (y + 7/2)^2 = 75/4
Now, we can differentiate both sides of the equation with respect to x using implicit differentiation. Note that since (x-2)^2 + (y+7/2)^2 = r^2, where r is the radius of the circle, differentiating both sides with respect to x gives:
2(x - 2) + 2(y + 7/2)(dy/dx) = 0
Simplifying this equation:
2x - 4 + (y + 7/2)(dy/dx) = 0
2x - 4 + (y + 7/2)(dy/dx) = 0
Now, we can solve for dy/dx:
(dy/dx) = (4 - 2x) / (y + 7/2)
For part B, to determine the conditions for the tangent line to be horizontal or vertical, we can analyze the derivative dy/dx. A tangent line is horizontal when the derivative is zero, and vertical when the derivative is undefined.
In our case, dy/dx becomes undefined when the denominator (y + 7/2) equals zero. Solving for y:
y + 7/2 = 0
y = -7/2
So, when y = -7/2, the tangent line is vertical.
Next, we need to find the values of x for which dy/dx equals zero. For that, we set the numerator equal to zero and solve for x:
4 - 2x = 0
2x = 4
x = 2
So, when x = 2, the tangent line is horizontal.
Therefore, the tangent line is horizontal when x = 2, and vertical when y = -7/2.
For part A, to find dy/dx, we can use implicit differentiation.
Start by differentiating both sides of the equation with respect to x:
d/dx (x^2 + y^2 - 4x + 7y) = d/dx (15)
The left side of the equation is the chain rule in action. Let's focus on differentiating each term one by one:
d/dx (x^2) = 2x
d/dx (y^2) = 2y * (dy/dx) (using the chain rule)
d/dx (-4x) = -4
d/dx (7y) = 7 * (dy/dx) (using the chain rule)
d/dx (15) = 0 (since it's a constant)
Now we can rewrite the equation:
2x + 2y * (dy/dx) - 4 + 7 * (dy/dx) = 0
Combine like terms:
(2x - 4) + (2y + 7) * (dy/dx) = 0
(2x - 4) + (2y + 7) * (dy/dx) = 0
Now, isolate (dy/dx) by moving the terms not involving (dy/dx) to the other side of the equation:
(2y + 7) * (dy/dx) = 4 - 2x
Divide both sides by (2y + 7):
(dy/dx) = (4 - 2x) / (2y + 7)
So, the derivative dy/dx is given by (4 - 2x) / (2y + 7).
For part B, to determine when the tangent line is horizontal or vertical, we need to analyze the slope of the tangent line, which is dy/dx.
For a tangent line to be horizontal, the slope dy/dx should be equal to 0. So, set the derivative to 0:
(4 - 2x) / (2y + 7) = 0
This equation is satisfied when the numerator is 0, i.e., 4 - 2x = 0. Solve for x:
4 - 2x = 0
2x = 4
x = 2
So, when x = 2, the tangent line to the curve is horizontal.
For a tangent line to be vertical, the slope dy/dx should be undefined or infinite. In this case, the denominator (2y + 7) should be 0. Solve for y:
2y + 7 = 0
2y = -7
y = -7/2
So, when y = -7/2, the tangent line to the curve is vertical.
To summarize:
- The tangent line is horizontal when x = 2.
- The tangent line is vertical when y = -7/2.