Problem 69 - Calculus Early Transcendentals

Question

Find an equation of the line tangent to the following curves at the given point. $$x^{2}=-6 y ;(-6,-6)$$

Step-by-Step Solution

Answer

Answer: The equation of the tangent line is $$y = 2x$$.

Step 1: Rewrite the given equation as a function of x and y

The given equation is already a function of x and y, so we can proceed to the next step without any changes.

Step 2: Rewrite the equation as a function of y

We need to rewrite the equation in the form of $$y = f(x)$$. By solving for y, we have: $$y = -\frac{1}{6}x^{2}$$

Step 3: Find the derivative dy/dx

To find the tangent line's slope at a given point, we need to calculate the derivative of y with respect to x: $$\frac{dy}{dx} = \frac{d}{dx} (-\frac{1}{6}x^{2})$$ Use the power rule: $$\frac{dy}{dx} = -\frac{1}{3}x$$

Step 4: Evaluate the derivative at the given point

Now, plug in the x-coordinate of the given point (-6) into the derivative to find the tangent line's slope at the point: $$m = -\frac{1}{3}(-6) = 2$$

Step 5: Use the point-slope form to find the equation of the tangent line

The point-slope form of a line's equation is given by: $$y - y_{1} = m (x - x_{1})$$ Plug in the coordinates of the given point and the slope calculated in step 4: $$y - (-6) = 2 (x - (-6))$$ Simplify: $$y + 6 = 2(x + 6)$$ Now, distribute and rearrange the equation to get the final equation of the tangent line: $$y = 2x + 6 - 6$$ $$y = 2x$$ The equation of the tangent line to the curve $$x^2 = -6y$$ at the point $$(-6, -6)$$ is $$y = 2x$$.