Chapter 7.4, Problem 105AYU

Searchlights A searchlight at the grand opening of a new car dealership casts a spot of light on a wall located 75 meters from the searchlight. The acceleration $\ddot{r}$ of the spot of light is found to be $\ddot{r}=1200\sec \theta \left(2{\sec }^2\theta -1\right)$. Show that this is equivalent to $\ddot{r}=1200\left(\frac{1+{\sin }^2\theta }{{\cos }^3\theta }\right)$.

To determine

To find: A searchlight at the grand opening of a new car dealership casts a spot of light on a wall located 75 meters from the searchlight. The acceleration $r$ of the spot of light is found to be $r=1200\sec \theta \left(2{\sec }^2\theta -1\right)$. Show that this is equivalent to $r=1200\left(\frac{1+{\sin }^2\theta }{{\cos }^3\theta }\right)$.

Answer to Problem 105AYU

Solution:

Equivalence between $r=1200\sec \theta \left(2{\text{sec}}^2\theta -1\right)\text{ and }r=1200\left(\frac{1+{\text{ sin}}^2\theta }{{\text{cos}}^3\theta }\right)\text{is found}$.

Explanation of Solution

Given:

$r=1200\sec \theta \left(2{\text{sec}}^2\theta -1\right)\text{and }r=1200(\frac{1+{\text{ sin}}^2\theta }{{\text{cos}}^3\theta })$

Calculation:

By reciprocal identity.

$r=1200\frac{1}{\cos \theta }\left(2\frac{1}{{\text{cos}}^2\theta }-1\right) \text{and }r=1200(\frac{1+{\text{ sin}}^2\theta }{{\text{cos}}^3\theta })$

Adding up the quotients.

$r=1200(\frac{2-{\cos }^2\theta }{{\cos }^3\theta })\text{ and }r=1200(\frac{1+{\text{ sin}}^2\theta }{{\text{cos}}^3\theta }\text{)}$.

$r=1200(\frac{1+1-{\cos }^2\theta }{{\cos }^3\theta })\text{ and }r=1200(\frac{1+{\text{ sin}}^2\theta }{{\text{cos}}^3\theta })$

By Pythagorean identity.

$r=1200(\frac{1+{\sin }^2\theta }{{\cos }^3\theta })\text{ and }r=1200(\frac{1+{\text{ sin}}^2\theta }{{\text{cos}}^3\theta })$

Both are equivalent.