Problem 53 - Essential Calculus Early Transcendentals

Question

Find a vector equation for the tangent line to the curve of intersection of the cylinders \(x^{2}+y^{2}=25\) and \(y^{2}+z^{2}=20\) at the point \((3,4,2)\)

Step-by-Step Solution

Answer

The vector equation is \(\mathbf{r}(t) = (3, 4, 2) + t(12, 16, 8)\).

Step 1: Parametrize the First Cylinder

The first cylinder is given by the equation \(x^2 + y^2 = 25\). We can use a trigonometric parametrization where \(x = 5\cos(t)\) and \(y = 5\sin(t)\). This represents the circular cross-section of the cylinder in the xy-plane with a radius of 5.

Step 2: Parametrize the Second Cylinder

The second cylinder is given by the equation \(y^2 + z^2 = 20\). Similarly, we can parametrize this as \(y = \sqrt{20}\cos(s)\) and \(z = \sqrt{20}\sin(s)\). This describes a circular cross-section in the yz-plane with a radius of \(\sqrt{20}\).

Step 3: Find the Tangent Vectors

We need the tangent vectors of both cylinders at the point \((3,4,2)\). For the first cylinder parametrized by \(x = 5\cos(t)\),\( y = 5\sin(t)\): The point condition \((3,4)\) suggests \((3 = 5\cos(t), 4 = 5\sin(t))\). The derivatives give \(\frac{dx}{dt} = -5\sin(t)\) and \(\frac{dy}{dt} = 5\cos(t)\). Similarly, for the second cylinder: \( (y,z) = (4,2)\rightarrow (4 = \sqrt{20}\cos(s), 2 = \sqrt{20}\sin(s))\). Derivatives are \(\frac{dy}{ds} = -\sqrt{20}\sin(s)\) and \(\frac{dz}{ds} = \sqrt{20}\cos(s)\).

Step 4: Evaluate Derivatives at the Given Point

Solve for \(t\) and \(s\) where \(x=3, y=4, z=2\). For the first, \(\cos(t) = \frac{3}{5}, \sin(t) = \frac{4}{5}\). For the second, \(\cos(s) = \frac{4}{\sqrt{20}}, \sin(s) = \frac{2}{\sqrt{20}}\). Insert into derivatives to find tangent vectors: For first cylinder: \(\left(-5\sin(t), 5\cos(t), 0\right) = \left(-4, 3, 0\right)\). For second cylinder: \(\left(0, -\sqrt{20}\sin(s), \sqrt{20}\cos(s)\right) = \left(0, -2, 4\right)\).

Step 5: Find the Tangent Line

The direction vector of the tangent line at the intersection point is the cross product of tangent vectors from both cylinders. Perform the cross product: \((-4, 3, 0) \times (0, -2, 4) = (12, 16, 8)\).So, the tangent line direction vector is \((12, 16, 8)\).

Step 6: Write the Vector Equation of the Tangent Line

Use the point \((3, 4, 2)\) and the direction vector \((12, 16, 8)\) to write the equation:\(\mathbf{r}(t) = (3, 4, 2) + t(12, 16, 8)\), where \(t\) is the parameter.

Key Concepts

Cylinder Intersections

The study of intersections of cylinders involves finding common points shared by the surfaces of two cylindrical structures. When addressing the intersection of two cylinders, such as those given by the equations \(x^2 + y^2 = 25\) and \(y^2 + z^2 = 20\), we are tasked with identifying the curve formed where these two surfaces overlap. Understanding intersections is crucial for problems involving spatial geometry.In this exercise, the intersection involves matching the geometry of a circle in the \(xy\)-plane with a radius of 5 and another circle in the \(yz\)-plane with a radius of \(\sqrt{20}\). These form a 3D curve where the two cylinders meet. Visualizing such intersections helps in grasping the spatial orientation and the resulting structure of curves or lines in higher dimensions.A common point on both cylindrical surfaces, such as \((3,4,2)\) in this case, lies on the curve of intersection. Identifying these key points assists in parametrizing the parts of the curves that overlap.

Parametrization

Parametrizing involves expressing a geometric figure using one or more parameters. For complex shapes like cylinders, parametrization makes it easier to describe and manipulate them mathematically.

• For the cylinder \(x^2 + y^2 = 25\), a circle with radius 5, the parametrization is given by \(x = 5\cos(t)\), \(y = 5\sin(t)\), where \(t\) is the parameter.
• For the cylinder \(y^2 + z^2 = 20\), a circle with radius \(\sqrt{20}\), it's \(y = \sqrt{20}\cos(s)\), \(z = \sqrt{20}\sin(s)\) with \(s\) as the parameter.

These parametric equations reduced the complexity from dealing with equations directly to dealing with trigonometric functions of a single variable. At a given point, these functions help determine the directional derivatives, which are essential in finding tangent vectors.

Cross Product

The cross product is used here to find a direction vector that is perpendicular to two given vectors. This is particularly useful in our exercise where determining the direction of the tangent line is necessary.To find the tangent line at the curve of intersection, we obtain two tangent vectors, one from each cylinder.

• First tangent vector: \((-4, 3, 0)\) derived from the parametric equations of the first cylinder.
• Second tangent vector: \((0, -2, 4)\) from the parametric representation of the second cylinder.

Performing the cross product of these vectors, \((-4, 3, 0) \times (0, -2, 4)\), provides us with the perpendicular direction \((12, 16, 8)\). This direction vector is crucial in forming the vector equation of the tangent line, reflecting how these mathematical operations are deployed to solve geometric problems.

Vector Equations

Vector equations are a handy tool for describing lines in space. They combine a point vector and a direction vector, parametrizing the path that extends from the given point in the direction of the vector. In the context of our solution, we derived the equation for a tangent line using the cross product result as a direction vector and a known point \((3, 4, 2)\) on the curve. The equation is structured as:\[\mathbf{r}(t) = (3, 4, 2) + t(12, 16, 8)\]Where \(\mathbf{r}(t)\) represents any point on the line, and \(t\) is a scalar parameter. The first part \((3, 4, 2)\) is the starting point, and \(t(12, 16, 8)\) describes moving in the direction of \((12, 16, 8)\). Vector equations like this are essential for identifying positions of points along a line and visualizing the tangential path intersecting the cylinder surfaces.