Derivatives (Trig) with Isosceles Triangles

[LadiesMan]

1. THe base of an isosceles triangle is 20 cm and the altitude is increasing at the rate of 1 cm/min. At whate rate is the base angle increasing when the area is 100 cm^2? answer 0.05 rad/s


2. What I did:
-Took the derivative of tan and assumed that each base side of the trianlge was 10 cm
-Found height by using area which also came up to be ten! h = 100/x (x is each base side)
-Sub values for the derivative equation and got 0.14 as an answer =(

Thanks

 

[Dick]

You are giving some of the correct ingredients to use in solving the problem, but you aren't giving us any clue how you put them together to get 0.14. Show us the actual equation you solved to get 0.14.

 

[lamerali]

The base of an isosceles triangle is 20 cm and the altitude is increasing at the rate of 1 cm/min. At what rate is the base angle increasing when the area is 100 cm2?

Hmmmm...I have the same question but i came up with a completely different answer...is this anywhere near correct?



The isosceles triangle divides into two right triangles with bases of 10 cm and areas 50 cm2

tan\theta = \frac{h}{10}

\frac{d\theta}{dt} sec^{2} \theta = \frac{1}{10}

\frac{d\theta}{dt} = \frac{1}{10}. cos ^{2} \theta

A = \frac{1}{2} b x h

100 = \frac{1}{2} (20) . h
h = 10

tan \theta = \frac{10}{10}
tan \theta = 1

we know,

sin^{2} \theta + cos^{2} \theta = 1

(cos\theta tan \theta) ^{2} + cos ^{2} \theta = 1

2 cos^{2} \theta = 1
cos^{2} \theta = \frac{1}{2}


\frac{d\theta}{dt} = \frac{1}{10} . cos^{2} \theta

= \frac{1}{10} . cos^{2}\frac{1}{2}
= 0.077 rads / min

or 0.0013 rads /s