Problem 2 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Which is larger, the slope of the tangent line to \(x^{2}+y^{2}=25\) at the point \((4,-3)\) or the slope of the tangent line to \(x^{2}+4 y^{2}=25\) at the point \((4,-3 / 2) ?\) You can answer this analytically (find the two slopes), or you can answer this by looking at the graphs of the ellipse and the circle and sketching the tangent to each at the designated points. (To sketch the ellipse, look at the \(x\) -and \(y\) -intercepts.)

Step-by-Step Solution

Answer

The slope of the tangent line to \(x^2 + y^2 = 25\) at the point \((4,-3)\) is larger.

Step 1: Differentiation of the given circle equation

We begin by implicitly differentiating the first equation \(x^2 + y^2 = 25\). Resulting in \(2x+2yy'=0\), we solve this equation for \(y'\) (the derivative of \(y\) with respect to \(x\), which gives us the slope of the tangent line at any point), \(y' = -x/y.\) Now we evaluate \(y'\) at the given point \((4,-3)\), to get the slope of the tangent line to the line at that point.

Step 2: Calculation of the slope at the given point on the circle

By substituting \(x = 4\) and \(y = -3\) in the equation \(y' = -x/y\), we get \(y' = -4/-3 = 4/3\). So, the slope of the tangent line to the circle at that point is \(4/3\).

Step 3: Differentiation of the given ellipse equation

We move on to the ellipse with equation \(x^2 + 4y^2 = 25\). Implicitly differentiate with respect to \(x\) to get \(2x + 8yy' = 0\). Solving for \(y'\), we find \(y' = -x/(4y)\). Then evaluate \(y'\) at the point \((4, -3/2)\).

Step 4: Calculation of the slope at the given point on the ellipse

Substituting \(x = 4\) and \(y = -3/2\) in the equation \(y' = -x/(4y)\), we get \(y' = -4/(-6) = 2/3\). Therefore, the slope of the tangent line to the ellipse at the given point is \(2/3\).

Step 5: Comparing the slopes of the tangent lines

From step 2, the slope of the tangent to the circle is \(4/3\). From step 4, the slope of the tangent to the ellipse is \(2/3\). Comparing these two, \(4/3 > 2/3\).

Key Concepts

Differentiation of Implicit Functions

Understanding the differentiation of implicit functions is crucial when dealing with equations that cannot be easily solved for one variable in terms of others. Implicit differentiation allows us to find the derivative of a function that is not explicitly defined. For example, when we have equations like a circle's equation, x^2 + y^2 = 25, where y is not isolated, we apply this method.

Here's how it works: we differentiate both sides of the equation with respect to x, and every time we come across y or a term involving y, we multiply by y' (the derivative of y with respect to x). This gives us an equation where we can solve for y', revealing the slope of the tangent line at a particular point.

Tangent Line to a Circle

The concept of a tangent line to a circle ties closely with implicit differentiation. A tangent line to a circle at a given point is the straight line that touches the circle at exactly one point. This line is also perpendicular to the radius of the circle at the point of tangency.

To find the slope of this tangent line, first, we implicitly differentiate the equation of the circle. In the given exercise, we differentiate x^2 + y^2 = 25 to find y', which is -x/y at the point (4, -3). Hence, we find that the slope is 4/3. This slope indicates the steepness and direction of the tangent line at that specific point on the circle.

Tangent Line to an Ellipse

Similar to a circle, an ellipse also has a tangent line at any of its points, and the process of finding this slope mirrors that of a circle using implicit differentiation. However, since the shape of an ellipse is elongated, the equation differs. It typically has the form ax^2 + by^2 = c.

For the provided ellipse x^2 + 4y^2 = 25, after differentiating and solving for y', we find the slope of the tangent line at the point (4, -3/2) to be 2/3. Notably, due to the ellipse's shape, the calculation includes a different factor that changes how the slope of the tangent is related to the point of tangency compared to a circle.

Comparing Slopes

When comparing slopes, we are essentially comparing the steepness of different lines. The slope is a measure of the line's inclination; a higher absolute slope value indicates a steeper line. In our exercise, we compared the slopes of tangent lines to a circle and an ellipse at specific points.

The slope of the tangent line to the circle at (4, -3) was found to be 4/3, while that of the ellipse at (4, -3/2) was 2/3. By comparing these values, we can see that 4/3 is larger, meaning the tangent line to the circle is steeper at the specified point. Understanding slope comparison is beneficial in different fields of mathematics and physics, particularly when analyzing the behavior of curves at given points.