Can someone show me how to solve it.

Triangle PQR is right angled at Q, and PQ is 6cm long. QR increases in length at 2cm per minute. Find the rate of change in QPR at the instant when QR is 8cm Long.

Thanks.

Guest Aug 15, 2017

#1**+2 **

"*Triangle PQR is right angled at Q, and PQ is 6cm long. QR increases in length at 2cm per minute. Find the rate of change in QPR at the instant when QR is 8cm Long*."

.

Alan
Aug 15, 2017

#2**+1 **

Thanks Alan,

My answer is the same. :)

I'm just practicing :)

\(\frac{dp}{dt}=2\;cm/min\\ \frac{d\theta}{dt}=\frac{d\theta}{dp}\cdot \frac{dp}{d\theta}\\ \frac{d\theta}{dt}=\frac{d\theta}{dp}\cdot2\\ \frac{d\theta}{dt}=2\frac{d\theta}{dp}\\ \frac{d\theta}{dt}=2\times\frac{1}{6sec^2\theta}\\ \frac{d\theta}{dt}=\frac{cos^2\theta}{3}\\ \frac{d\theta}{dt}=\frac{36}{3q^2}\qquad \text{from diagram}\\ \frac{d\theta}{dt}=\frac{12}{(36+p^2)}\\~\\ When\;\; p=8\\ \frac{d\theta}{dt}=\frac{12}{(36+64)}\\ \frac{d\theta}{dt}=0.12\; radians/min \)

Melody
Aug 15, 2017