Derivatives of trig functions and isosceles triangles.

lamerali
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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?

I wasnt really sure where to start on this question so i tried my best at an answer. I'm sure I've gone wrong with this question so i appreciate any guidance.

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

50 = \frac{1}{2} (10) . 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 4.6 rads /s


Thanks in advance!
 
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lamerali said:
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?

I wasnt really sure where to start on this question so i tried my best at an answer. I'm sure I've gone wrong with this question so i appreciate any guidance.

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

50 = \frac{1}{2} (10) . h
h = 10
Here's your error. "when the area is 100 cm2" refers to the original isosceles triangle and that had base 20, not 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 4.6 rads /s


Thanks in advance!
 
HallsofIvy said:
Here's your error. "when the area is 100 cm2" refers to the original isosceles triangle and that had base 20, not 10.

but if i plug in
A = (1/2) b x h

100 = (1/2) (20) h

h is still equal to 10 cm. Is that the only error you see? because that does not effect the rest of the equation...:S
 
is the rest of this equation anywhere near correct?? :(
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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