Calculus: Applications of Trig

[erik05]

Hello all. I have a test on these type of questions coming up soon :grumpy: and this is about the time when frustration sets in since I'm not very good at these questions. If anyone could show me how to do them so I could study off it then you would be my hero.

1) A wall of height 8 m stands parallel to and 27 m from a tall building. A ladder with its foot on the ground is to pass over the wall and lean on the building. What angle will the shortest such ladder make with the ground? Ans: 0.58800

I think to get to the right answer, you have to end up with tanx= 2/3 and take the inverse to get 0.58800. Not quite sure how to get to that though.

2) A kite 40 m above the ground moves horizontally at the rate of 3 m/s. At what rate is the angle between the string and the horizontal decreasing when 80 m of string has been let out? Ans: 0.02 m/s

So far all I have is a triangle with a=40, b=x, c= 80 so cosØ= x/80. Taking the derivative: dx/dt= 80-sinØ dØ/dt. This was where I got stuck.

3) Two sides of a triangle are 6 and 8 metres in length. If the angle between them decreases at the rate of 0.035 rad/s, find the rate at which the area is decreasing when the angle between the sides of fixed length is $$\frac {\pi}{6}$$ ? Ans:0.727 m^2/min

Would the cosine law be involved in this to find dA/dt?

As you can see from my feeble attempts, I really suck at this. Any help with how to get the solutions would be much appreaciated. Thanks.

 

[Jameson]

For optimization problems, the hardest part is coming up with an equation to work with in the first place. For the first question, what kind of shape are you dealing with and what equation could be used to describe it? Same thing for the other two. Show what work you've done so far and what eqation(s) you think you might need. For the third question, is the shape a right triangle or some other triangle?

Jameson

 

[erik05]

That's the thing, I never know what to start with so I really don't have that much work. I'm assuming that all of these questions are involving triangles and that the cosine law could come in handy for a few of them.

 

[whozum]

2) A kite 40 m above the ground moves horizontally at the rate of 3 m/s. At what rate is the angle between the string and the horizontal decreasing when 80 m of string has been let out? Ans: 0.02 m/s

So far all I have is a triangle with a=40, b=x, c= 80 so cosØ= x/80. Taking the derivative: dx/dt= 80-sinØ dØ/dt. This was where I got stuck.

I was goingthrough this and got stuck too, but maybe one of my mistakes will help you out, I got 0.075
Alright, write down everything you know,

a = hieght = 40m, constant
b = c*cos(u) = 80cos(u)
db/dt = 3
c = hypotenuse = string length = 80m
u = angle

What relates these?

$$\cos{u} = \frac{b}{c}$$

$$b = c*\cos{u}$$

$$\frac{db}{dt} = -c\sin{u}\frac{du}{dt}$$

$$\frac{du}{dt} = \frac{\frac{db}{dt}}{-c\sin{u}}$$

$$\sin{u} = \frac{a}{c} = 0.5$$

$$\frac{du}{dt} = \frac{3}{-(80)(0.5)} = 0.075$$

 

[erik05]

I get the same answer too except that it's negative, which makes sense since it's decreasing. This is what I did:

$$sin\theta= \frac {40}{80}$$ and taking the inverse of that I get 30 degrees. From before:

$$\frac {dx}{dt}= -80sin\theta \frac {d\theta}{dt}$$

$$\frac {d\theta}{dt}= \frac {\frac {dx}{dt}}{-80sin\theta}$$

$$\frac {3}{-80sin(30)} = -0.075$$


Hmmm...interesting.

 

[whozum]

How is the answer in meters per second?

 

[erik05]

That's what the answer in the back of the book says. It's weird because you'd think it would be in rad/s or degrees/s...

 

[whozum]

That' can't be right..

 

[erik05]

I guess I'll just assume that the book is wrong.

 

[whozum]

Give the others a shot.

 

[jai6638]

wwhats the answer to the first one?

from the diagram i drew, there are two triangles. I first found the length of the hypotenuse ( which is common for both triangles ) and then found the archcosine of 27/28 ( where 28 m is the length of the hypotenuse) which gave me 8 degrees..

Damn.. i don't know where my calculations are giong.. not gettin an answer.. can anyone help?