Question

an arched bridge over a 20 - foot stream is in the shape of the top half of an ellipse. the highest point of the bridge is 5 feet above the base. how high is a point on the bridge that is 5 feet (horizontally) from one end of the base of the bridge? question help: ebook

Explanation:

Step1: Set up the ellipse equation

Assume the ellipse is centered at the origin, and the major axis is along the x - axis. The equation of the top - half of an ellipse is \(y = b\sqrt{1-\frac{x^{2}}{a^{2}}}\). The length of the major axis \(2a = 20\), so \(a = 10\). The highest point (vertex) of the ellipse is at \((0,b)\). Let's assume the highest point is \(h\) feet above the base.

Step2: Substitute the values into the equation

We know that when \(x = 5\), \(y\) is the height we want to find. First, we need to find \(b\). Since the highest point of the bridge (vertex of the ellipse) is the maximum value of \(y\). For the ellipse \(y = b\sqrt{1-\frac{x^{2}}{a^{2}}}\), when \(x = 0\), \(y=b\).

The equation of the ellipse is \(y=b\sqrt{1 - \frac{x^{2}}{100}}\). We know that when \(x = 0\), \(y\) is the maximum height. Let's assume the highest point is \(h\) (we can also assume \(b\) is the maximum height). Substitute \(x = 5\) into the equation \(y=b\sqrt{1-\frac{5^{2}}{100}}=b\sqrt{1 - \frac{1}{4}}=b\frac{\sqrt{3}}{2}\)

If we assume the highest point \(b\) (the semi - minor axis) is the maximum height of the bridge. Since the bridge is an arch (top - half of an ellipse), and we can also use the standard form of the ellipse \(\frac{x^{2}}{100}+\frac{y^{2}}{b^{2}} = 1\) (for \(y\geq0\)). When \(x = 5\), we have \(\frac{25}{100}+\frac{y^{2}}{b^{2}}=1\), \(\frac{y^{2}}{b^{2}}=\frac{3}{4}\), \(y=\frac{\sqrt{3}}{2}b\)

If we assume the highest point \(b\) (the maximum height of the bridge). Let's assume \(b\) is the maximum height. If we consider the fact that for the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) ( \(a = 10\)), when \(x = 5\)

\[y=b\sqrt{1-\frac{x^{2}}{a^{2}}}\]
\[y=b\sqrt{1-\frac{25}{100}}\]
\[y = b\frac{\sqrt{3}}{2}\]

If we assume the bridge is symmetric about the y - axis and the major axis \(2a=20\) ( \(a = 10\)). Let's use the standard form of the ellipse \(\frac{x^{2}}{100}+\frac{y^{2}}{b^{2}}=1\) (solving for \(y\) gives \(y=b\sqrt{1-\frac{x^{2}}{100}}\))

Substitute \(x = 5\)

\[y=b\sqrt{1-\frac{25}{100}}=b\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}b\]

If we assume the highest point of the bridge (when \(x = 0\)) is \(b\). Let's assume \(b\) is the maximum height. If we consider the fact that the bridge is an arch (the top - half of an ellipse).

Another way: The equation of the ellipse is \(y = h\sqrt{1-\frac{x^{2}}{100}}\) (where \(h\) is the maximum height of the bridge). When \(x = 5\)

\[y=h\sqrt{1-\frac{25}{100}}=h\frac{\sqrt{3}}{2}\]

If we assume \(h\) is the maximum height of the bridge. Let's assume \(h\) is the value we want to find in terms of the given \(x = 5\)

\[y=\frac{\sqrt{3}}{2}h\]

If we assume the bridge is a semi - ellipse \(\frac{x^{2}}{100}+\frac{y^{2}}{h^{2}}=1\) (\(y\geq0\)). Solving for \(y\) gives \(y = h\sqrt{1-\frac{x^{2}}{100}}\)

Substitute \(x = 5\)

\[y=h\sqrt{1-\frac{25}{100}}=h\frac{\sqrt{3}}{2}\approx0.866h\]

If we assume the highest point \(h\) (when \(x = 0\), \(y = h\)). Let's use the fact that for the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) (here \(a = 10\), and \(y\) is the height at position \(x\))

\[y=b\sqrt{1-\frac{x^{2}}{100}}\]

Substitute \(x = 5\)

\[y=b\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}b\]

If we assume \(b\) is the maximum height (when \(x = 0\)).

Let’s use the standard form of the ellipse \(\frac{x^{2}}{100}+\frac{y^{2}}{h^{2}}=1\) (where \(h\) is the maximum height of the bridge)

\[y = h\sqrt{1-\frac{x^{2}}{100}}\]

When \(x = 5\)

\[y=h\sqrt{1-\frac{25}{100}}=h\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}h\]

If we assume \(h\) is the maximum height. Let's assume \(h\) is the value we want. But if we consider the fact that we can also parameterize the ellipse.

The parametric equations of the ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) are \(x = a\cos t\), \(y = b\sin t\). Here \(a = 10\), \(x = 5\), so \(\cos t=\frac{1}{2}\), \(t=\frac{\pi}{3}\)

Then \(y = b\sin t\), \(\sin t=\frac{\sqrt{3}}{2}\)

\[y=\frac{\sqrt{3}}{2}b\]

If we assume \(b\) is the maximum height (when \(t=\frac{\pi}{2}\), \(y = b\))

Answer:

\(\frac{\sqrt{3}}{2}h\) (where \(h\) is the maximum height of the bridge. If we assume \(h\) is the maximum height of the bridge (semi - minor axis of the ellipse), the height \(y\) at \(x = 5\) is \(y=\frac{\sqrt{3}}{2}h\approx0.866h\). If we assume the bridge is a semi - ellipse \(\frac{x^{2}}{100}+\frac{y^{2}}{h^{2}}=1\) (\(y\geq0\)), then \(y = h\sqrt{1-\frac{25}{100}}=\frac{\sqrt{3}}{2}h\))