Hey all, need help on the attached problems!
For the first problem, #9, imagine a right triangle consisting of the line from the light perpendicular to the wall, the line from the light to the dot of light on the wall, and the line on the wall from that dot of light to the base of the perpendicular. The angle in the right triangle at the light is $\displaystyle \theta$. The length of the side perpendicular to the wall is 11 miles. The distance from the dot of light to the base of the perpendicular is $\displaystyle x= 11 tan(\theta)$
So what is $\displaystyle \frac{dx}{dt}$ in terms of $\displaystyle \frac{d\theta}{dt}$?
2 "revolutions per minute" is $\displaystyle 2\pi$ radians per minute so $\displaystyle 120\pi$ radians per hour. $\displaystyle \frac{d\theta}{dt}= 120\pi$ radians per hour.
Now, try the others and post your work here.
For the first problem, #9, imagine a right triangle consisting of the line from the light perpendicular to the wall, the line from the light to the point on the wall, and the line on the wall from that dot of light to the base of the perpendicular. The angle in the right triangle at the light is $\displaystyle \theta$. The length of the side perpendicular to the wall is 11 miles. The distance from the dot of light to the base of the perpendicular is $\displaystyle x= 11 tan(\theta)$
So what is $\displaystyle \frac{dx}{dt}$ in terms of $\displaystyle \frac{d\theta}{dt}$?
2 "revolutions per minute" is [tex]2(2\pi)= 4\pi[tex] radians per minute so $\displaystyle 60(4\pi)= 240\pi$ radians per hour. $\displaystyle \frac{d\theta}{dt}= 240\pi$ radians per hour.
A helpful strategy I've found with related rates is to conjure a relationship between two variables, take the derivative with respect to time and you're done. The only thing left to do is to substitute the rate of change one of the quantitys is changing at.