Derivatives of trig functions and isosceles triangles.

  • Thread starter 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[tex]\theta[/tex] = [tex]\frac{h}{10}[/tex]

[tex]\frac{d\theta}{dt}[/tex] sec[tex]^{2}[/tex] [tex]\theta[/tex] = [tex]\frac{1}{10}[/tex]

[tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{1}{10}[/tex]. cos [tex]^{2}[/tex] [tex]\theta[/tex]

A = [tex]\frac{1}{2}[/tex] b x h

50 = [tex]\frac{1}{2}[/tex] (10) . h
h = 10

tan [tex]\theta[/tex] = [tex]\frac{10}{10}[/tex]
tan [tex]\theta[/tex] = 1

we know,

sin[tex]^{2}[/tex] [tex]\theta[/tex] + cos[tex]^{2}[/tex] [tex]\theta[/tex] = 1

(cos[tex]\theta[/tex] tan [tex]\theta[/tex]) [tex]^{2}[/tex] + cos [tex]^{2}[/tex] [tex]\theta[/tex] = 1

2 cos[tex]^{2}[/tex] [tex]\theta[/tex] = 1
cos[tex]^{2}[/tex] [tex]\theta[/tex] = [tex]\frac{1}{2}[/tex]


[tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{1}{10}[/tex] . cos[tex]^{2}[/tex] [tex]\theta[/tex]

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

or 4.6 rads /s


Thanks in advance!
 

Answers and Replies

  • #2
HallsofIvy
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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[tex]\theta[/tex] = [tex]\frac{h}{10}[/tex]

[tex]\frac{d\theta}{dt}[/tex] sec[tex]^{2}[/tex] [tex]\theta[/tex] = [tex]\frac{1}{10}[/tex]

[tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{1}{10}[/tex]. cos [tex]^{2}[/tex] [tex]\theta[/tex]

A = [tex]\frac{1}{2}[/tex] b x h

50 = [tex]\frac{1}{2}[/tex] (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 [tex]\theta[/tex] = [tex]\frac{10}{10}[/tex]
tan [tex]\theta[/tex] = 1

we know,

sin[tex]^{2}[/tex] [tex]\theta[/tex] + cos[tex]^{2}[/tex] [tex]\theta[/tex] = 1

(cos[tex]\theta[/tex] tan [tex]\theta[/tex]) [tex]^{2}[/tex] + cos [tex]^{2}[/tex] [tex]\theta[/tex] = 1

2 cos[tex]^{2}[/tex] [tex]\theta[/tex] = 1
cos[tex]^{2}[/tex] [tex]\theta[/tex] = [tex]\frac{1}{2}[/tex]


[tex]\frac{d\theta}{dt}[/tex] = [tex]\frac{1}{10}[/tex] . cos[tex]^{2}[/tex] [tex]\theta[/tex]

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

or 4.6 rads /s


Thanks in advance!
 
  • #3
62
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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
 
  • #4
62
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is the rest of this equation anywhere near correct?? :(
 

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