Answer to Question #156324 in Calculus for Phyroe

Question #156324

The base of an isosceles triangle 10 feet long and the base angles are decreasing at a rate of 2 degress per second. Find the rate of change of the area when the base angles are 45 degrees?


1
Expert's answer
2021-01-19T18:08:15-0500


Here in the figure BC=10ft. And we assume B=C=θ\angle B=\angle C =\theta (cause isosceles triangle).


Also BD=DC=5ft (cause AD is the perpendicular bisector of BC).


Now, in ABD\triangle ABD , using trigonometry:-



tan(ABD)=ADBD tanθ=AD5 AD=5tanθ\tan(\angle ABD)=\frac{AD}{BD}\\~\\ \Rightarrow\tan\theta=\frac{AD}{5}\\~\\ \Rightarrow AD=5\tan\theta

So, now we calculate the area of the ABC\triangle ABC :-



=125tanθ10 =25tanθ\triangle=\frac{1}{2}\cdot5\tan\theta\cdot10\\~\\ \Rightarrow\triangle=25\tan\theta\\

Now, as we're dealing with rate of change of different stuff we differentiate both sides wrt tt :-



ddt=25sec2θdθdt ddt=25(2)22 ddt=100\frac{d\triangle}{dt}=25\sec^2\theta\frac{d\theta}{dt}\\~\\ \Rightarrow \frac{d\triangle}{dt}=25\cdot(\sqrt{2})^2\cdot2\\~\\ \Rightarrow \frac{d\triangle}{dt}=100

We, put θ=45°\theta=45^\degree , and use the rate of change of θ\theta given in the question.


So, rate of change of area is 100100 sq ft per second.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment