Derivatives of trig functions and isosceles triangles.

[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?

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!

 

[HallsofIvy]

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!

 

[lamerali]

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

 

[lamerali]

is the rest of this equation anywhere near correct?? :(