Question

a. Determine an equation of the tangent line and normal line at the given point \(\left(x_{0}, y_{0}\right)\) on the following curves. b. Graph the tangent and normal lines on the given graph. $$\begin{aligned} &3 x^{3}+7 y^{3}=10 y\\\ &\left(x_{0}, y_{0}\right)=(1,1) \end{aligned}$$

Short answer

Question: Find the equations of the tangent and normal lines to the curve defined by \(3x^3 + 7y^3 - 10y = 0\) at the point (1, 1) and graph them. Answer: The equation of the tangent line is \(y=x\), and the equation of the normal line is \(y=-x+2\). Graph the tangent and normal lines by plotting the equations: 1. Tangent Line equation: \(y=x\) 2. Normal Line equation: \(y=-x+2\)

Step-by-step solution

Differentiating the given function with respect to x

To find the slope of the tangent line, we need to find the derivative of the given function, with respect to x. Following the implicit differentiation technique: $$\frac{d}{d x}(3 x^{3}+7 y^{3}-10 y) = \frac{d}{d x}(3 x^{3})+\frac{d}{d x}(7 y^{3})-\frac{d}{d x}(10 y)=0$$

Simplifying the derivative

Now, we need to find dy/dx by simplifying the derivative: $$9 x^{2}+21 y^{2}\frac{d y}{d x}-10\frac{d y}{d x}=0$$ $$9(1)^{2}+21(1)^{2}\frac{d y}{d x}-10\frac{d y}{d x}=0$$ $$9+21\frac{d y}{d x}-10\frac{d y}{d x}=0$$ $$(21-10)\frac{d y}{d x}=9$$ $$\frac{d y}{d x}=1$$ Thus, the value of \(\frac{d y}{d x}\) or the slope (m) is 1.

Getting the equation for the tangent line

Using the point-slope form of the line, we can get the equation of the tangent line: $$y - y_{0} = m (x - x_{0})$$ $$y - 1 = 1 (x - 1)$$ $$y=x$$ This is the equation of the tangent line.

Getting the equation for the normal line

Since the normal line is perpendicular to the tangent line, its slope will be the negative reciprocal of the tangent line's slope, m = -1. Now, using the point-slope form of the line again, we can get the equation of the normal line: $$y - y_{0} = m (x - x_{0})$$ $$y - 1 = -1 (x - 1)$$ $$y=-x+2$$ This is the equation of the normal line.

Graph the tangent and normal lines

To graph the lines, simply plot the equations we found in steps 3 and 4: 1- Tangent Line equation: \(y=x\) 2- Normal Line equation: \(y=-x+2\)

Key concepts

Tangent Line Equation

The equation of a tangent line reflects the slope and position of the line that just grazes a curve at a particular point without crossing it. To find this equation, we first need the derivative of the function to ascertain the slope at the point of tangency. Implicit differentiation comes to play when the function isn't explicitly solved for one variable.
In our example, given the curve 3x^3 + 7y^3 = 10y, we determined the derivative implicitly and found the slope to be 1 at the point (1,1). Using the point-slope form, y - y_0 = m(x - x_0), with m = 1 and (x_0, y_0) = (1,1), we got the equation of the tangent line, y = x. This tells us that at the point (1,1), the curve has a gradient of 1, and the line that touches the curve without crossing has the same gradient.

Normal Line Equation

While a tangent line touches a curve at a given point, the normal line is perpendicular to the curve at that same point. The slope of the normal line is the negative reciprocal of the tangent line’s slope. For our problem, since the slope of the tangent line is 1, the normal line’s slope will be -1 (the negative reciprocal of 1).
Using the same point-slope form but with m = -1 and the same point (1,1), we deduced the equation y = -x + 2. This gives us the path of the normal line, which is a straight line that crosses the curve at (1,1) at a right angle to the tangent.

Graphing Lines

Graphing lines effectively means visualizing mathematical relationships on a coordinate plane. Our problem required plotting the tangent and normal lines on a graph. For the tangent line y = x, every point where the x and y coordinates are equal will be on the line. It is a straight line at a 45-degree angle to both the x and y axes.
The normal line y = -x + 2 has a slope of -1, which means for every 1 unit increase in x, y decreases by 1 unit. To graph this line, we start at the y-intercept (where the line crosses the y-axis) at y = 2 and follow the slope downwards to the right. Importantly, for plotting both lines, make sure they intersect at the point (1,1), confirming their relationship to the original curve.

Derivative

A derivative represents how a function changes as its input changes—essentially, it’s the rate of change or the slope of the curve at any given point. In calculus, finding the derivative of a function is fundamental and serves multiple purposes such as determining the maximum and minimum values of functions, and understanding the rate of change in physical phenomena. In our exercise, we used the derivative, obtained through implicit differentiation, to identify the slope at a particular point on the curve. Knowing that the derivative is 1 at the point (1,1), allowed us to construct the precise equations for the tangent and normal lines.