EXAMPLE 1 (a) If x2 + y2- 25, find dy dx (b) Find an equation of the tangent to the circle x2 + y2 – 25 at the point (3, 4). SOLUTION 1 (a) Differentiating both sides of the equation x2+y25 )-) (25) + dx dx d 2x+2y X + dx 0. dx Remembering that y is a function of x and using the Chain Rule, we have 2y dy dx -2x X dx dx Thus 2y 2x + = 0. dx Now we solve this equation for dy/dx dy dx X y (b) At the point (3, 4) we have x = 3 and y= 4, so dy -3/4 dx An equation of the tangent to the circle at (3, 4) is therefore 4 -3/4 (x y- or 3 25 y X 4 SOLUTION 2 (b) Solving the equation x2y-25, we get y t25-x2. The point (3, 4) lies on the upper semicircle y 25-x and so we consider the function fx)- 25-Differentiating f using the Chain Rule, + we have 125–/2 V25- f(x) dx Your answer is incorrect. as- (25-2x) x ) So f(3) and, as in Solution 1, an equation of the tangent is
The Correct Answer and Explanation is :
Solution to the Given Problem
(a) Finding ( \frac{dy}{dx} )
We start with the given equation of the circle:
[
x^2 + y^2 = 25
]
Differentiating both sides with respect to ( x ), using implicit differentiation:
[
\frac{d}{dx} (x^2) + \frac{d}{dx} (y^2) = \frac{d}{dx} (25)
]
[
2x + 2y \frac{dy}{dx} = 0
]
Solving for ( \frac{dy}{dx} ):
[
2y \frac{dy}{dx} = -2x
]
[
\frac{dy}{dx} = \frac{-x}{y}
]
(b) Finding the Equation of the Tangent Line
At the given point ( (3,4) ), we substitute ( x = 3 ) and ( y = 4 ):
[
\frac{dy}{dx} = \frac{-3}{4}
]
The equation of a tangent line at ( (x_1, y_1) ) with slope ( m ) is:
[
y – y_1 = m(x – x_1)
]
Substituting ( (3,4) ) and ( m = -\frac{3}{4} ):
[
y – 4 = -\frac{3}{4} (x – 3)
]
Expanding:
[
y – 4 = -\frac{3}{4} x + \frac{9}{4}
]
[
y = -\frac{3}{4} x + \frac{9}{4} + 4
]
[
y = -\frac{3}{4} x + \frac{25}{4}
]
Thus, the equation of the tangent line is:
[
y = -\frac{3}{4}x + \frac{25}{4}
]
Explanation (300 Words)
A circle is defined as the set of all points equidistant from a fixed center. The given equation, ( x^2 + y^2 = 25 ), represents a circle centered at the origin ( (0,0) ) with a radius of ( 5 ) since ( 25 = 5^2 ).
To find the tangent line, we first compute the derivative ( \frac{dy}{dx} ) using implicit differentiation, which is useful when ( y ) is a function of ( x ) but cannot be explicitly solved for easily. Differentiating both sides, we apply the chain rule to ( y^2 ), which gives ( 2y \frac{dy}{dx} ). Solving for ( \frac{dy}{dx} ), we obtain ( -\frac{x}{y} ).
At the specific point ( (3,4) ), substituting ( x = 3 ) and ( y = 4 ) into the derivative formula yields ( \frac{dy}{dx} = -\frac{3}{4} ).
The tangent line follows the equation ( y – y_1 = m(x – x_1) ), where ( (x_1, y_1) = (3,4) ) and ( m = -\frac{3}{4} ). Substituting these values and simplifying, we obtain the final equation:
[
y = -\frac{3}{4} x + \frac{25}{4}
]
This line represents the tangent to the circle at ( (3,4) ), meaning it touches the circle at exactly that point without crossing it. This solution confirms that the previous answer was incorrect due to differentiation mistakes.