Re: Related Rates - Angles

Quote:

Originally Posted by

**rofredtan** **Note:** I already got part a, but posted it because it includes information useful for part b.

A baseball diamond is a square 90 ft on a side. A player runs from first base to second at a rate of 16 ft/s.

a)At what rate is the player's distance from third base changing when the player is 30 ft from first base?

b) At what rates are angles theta1 and theta2 (see the figure) changing at that time?

http://img62.imageshack.us/img62/3462/2ivgwzo.jpg
My try:

http://img13.imageshack.us/img13/470/spg8kh.png
The

**correct answers for part b** are 8/65 rad/sec (theta1) and -8/65 rad/sec (theta2)

Thanks in advance,

rofredtan

Hi rofredtan,

You have used the product rule incorrectly.

If u and v are two differentiable functions of t; then,

$\displaystyle \frac{d(uv)}{dt}=u\frac{dv}{dt}+v\frac{du}{dt}$

In your case let, $\displaystyle u=\frac{1}{f}\mbox{ and }v=z$

$\displaystyle \tan\theta_{1}=\frac{z}{f}\Rightarrow \sec^{2}\theta_{1}\frac{d\theta_{1}}{dt}=\frac{1}{ f}\frac{dz}{dt}-\frac{1}{f^2}\frac{dz}{dt}$

Hope you can continue.

Re: Related Rates - Angles

Also your picture is wrong. You show the distance from the runner to third base as 30' when the problem says it 30' from the runner to **first** base.

Re: Related Rates - Angles

Hello, rofredtan!

Quote:

$\displaystyle \text{A baseball diamond is a square 90 ft on a side.}$

$\displaystyle \text{A player runs from first base to second at a rate of 16 ft/s.}$

$\displaystyle \text{a) At what rate is the player's distance from third base changing}$

. . $\displaystyle \text{when the player is 30 ft from first base?}$

$\displaystyle \text{b) At what rates are angles }\alpha\text{ and }\beta\text{ changing at that time?}$ Code:

` C`

* x

* * A

90 * α o

* o *

* β o *

* o *

B o *

* *

* *

* *

* *

* *

*

The player is at A; third base is B; second base is C.

$\displaystyle \text{Let }x\text{ = distance }AC.\; \text{ Note that: }\,\frac{dx}{dt}\,=\,-16\text{ ft/s}$

$\displaystyle \text{Let }\alpha = \angle CAB,\;\beta = \angle CBA$

$\displaystyle \text{We have: }\:\tan\alpha \,=\,\frac{90}{x} \quad\Rightarrow\quad \alpha \:=\:\tan^{-1}\left(\frac{90}{x}\right)$

$\displaystyle \text{Then: }\:\frac{d\alpha}{dt} \:=\:\frac{1}{1 + (\frac{90}{x})^2}\,\left(-\frac{90}{x^2}\right)\left(\frac{dx}{dt}\right)$

$\displaystyle \text{When }x = 60\!:\;\frac{d\alpha}{dt} \;=\;\frac{1}{1 + (\frac{90}{60})^2}\left(-\frac{90}{60^2}\right)(-16) $

. . $\displaystyle \frac{d\alpha}{dt} \;=\;\frac{1}{1 + \frac{4}{9}}\left(-\frac{1}{40}\right)(-16) \;=\;\frac{1}{\frac{13}{4}}\left(\frac{2}{5}\right ) \;=\;\frac{4}{13}\cdot\frac{2}{5}$

$\displaystyle \text{Therefore: }\:\frac{d\alpha}{dt} \:=\:\frac{8}{65}\text{ rad/sec.} $

$\displaystyle \text{Since }\,\beta \:=\:\frac{\pi}{2} - \alpha,\,\text{ then: }\:\frac{d\beta}{dt} \:=\:-\frac{d\alpha}{dt}$

$\displaystyle \text{Therefore: }\:\frac{d\beta}{dt} \:=\:-\frac{8}{65}\text{ rad/sec.}$