A Ton of Calc Questions

[jordan]

Yes, this is homework, but it's under the calc section because I'm looking for a discussion more than just quick answers.

 

1) a) Find the everage temperature, to the nearest degree, between t=6 and t=14 for all [math]F(t)=80-10\cos(\frac{\pi t}{12})[/math]

 

[math]\frac{1}{8}\int_{6}^{14}80-10\cos(\frac{\pi t}{12})dt[/math]

 

[math]\frac{1}{8}[80t+\frac{10\pi}{12}\sin(\frac{\pi t}{12})]_{6}^{14}[/math]

 

[math]\frac{5}{4}[8t+\frac{\pi}{12}\sin(\frac{\pi t}{12})]_{6}^{14}[/math]

 

[math]\frac{5}{4}[8(14)+\frac{\pi}{12}\sin(\frac{7\pi}{6})]-\frac{5}{4}[8(6)+\frac{\pi}{12}\sin(\frac{\pi}{2})][/math]

 

This yeilds 79.5 or so. The only problem [math]80 \leq F(t) \leq 90[/math], so the average can't be in the 70s. Can anyone find where the mistake is?

 

b)An air conditioner cooled the house whenever the outside temperature was at or above 78 degrees. For what values of t was the air conditioner on?

 

[math]78=80-10\sin(\frac{\pi t}{12})[/math]

 

[math].2=\sin(\frac{\pi t}{12})[/math]

 

[math]\frac{12\arcsin(.2)}{\pi}=t[/math]

 

That gives me .7 something, which again is wrong acording to the graph. Plus, I should have 2 answers. Any clues on where this one went wrong?

 

c)Cost of cooling this house is $0.05 per hour for each degree over 78. What was the total cost for cooling the house for one 24 hour period?

 

I plan to get the answers from part b and use them as the bounds when I integrate the function. That would give me the area under the curve for the tiime the air conditioner was on. Then multiply that by .05 and that should give the cost, right?

 

 

 

2) a) A curve is defined by 2y³+6x²y-12x²+6y=1. Show that [math]\frac{dy}{dx}=\frac{4x-2xy}{x^2+y^2+1}[/math]

 

I got that part.

 

b) Write an equation for each horizontal tangent line to the curve.

 

A horizontal tangent line would be were the second derivitive would be 0, but taking the second derivitive of that seems insane. Is there an easier way to do this one?

 

 

Also, if there are any easier/quicker meathods to any previous problems could you also point those out. Thanks.

[jordan]

Also, can anyone tell me what [math]\oint[/math] is called and/or means?

[Dave]

This yeilds 79.5 or so. The only problem [math]80 \leq F(t) \leq 90[/math'], so the average can't be in the 70s. Can anyone find where the mistake is?

 

Nope, [math]70 \leq F(t) \leq 90[/math] (80 -10 doesn't equal 80 ).

 

That gives me .7 something, which again is wrong acording to the graph. Plus, I should have 2 answers. Any clues on where this one went wrong?

 

Remember that sin is periodic. arcsin will give you one answer because to be a well-defined function it can only give you one answer. Once you find one value, it's easy to find the next value by looking at the symmetry of the graph. Also, did you have your calculator on degrees mode?

 

c)Cost of cooling this house is $0.05 per hour for each degree over 78. What was the total cost for cooling the house for one 24 hour period?

 

I plan to get the answers from part b and use them as the bounds when I integrate the function. That would give me the area under the curve for the tiime the air conditioner was on. Then multiply that by .05 and that should give the cost, right?

 

That sounds about right to me.

 

b) Write an equation for each horizontal tangent line to the curve.

 

A horizontal tangent line would be were the second derivitive would be 0, but taking the second derivitive of that seems insane. Is there an easier way to do this one?

 

Think about this one for a second: the derivative will give you the gradient of the curve. If you have a horizontal tangent, what does this imply?

[Dave]

Also, can anyone tell me what [math]\oint[/math'] is called and/or means?

 

It's a closed integral. Basically, you can integrate functions along paths in the plane. Think of a wire in three dimensions; you might have some function that tells you the density of the wire at some particular point. To find the total density of the wire, you'd have to do a line integral along the length of the wire. When the two ends of the wire are connected to each other, you're integrating around a closed loop and you can use that symbol.

[jordan]

Nope, [math]70 \leq F(t) \leq 90[/math'] (80 -10 doesn't equal 80 ).

 

But on the interval of 6-14 the funtion ranges between 80 and 90.

 

Also, did you have your calculator on degrees mode?

Nope.

 

Think about this one for a second: the derivative will give you the gradient of the curve. If you have a horizontal tangent, what does this imply?

So I only need to set the first derivitive equal to zero?

[Dave]

But on the interval of 6-14 the funtion ranges between 80 and 90.

 

There is that. However, plugging the integral into mathematica tells me the answer is 87.162, so you may have evaluated it wrongly; the working looks okay though.

 

So I only need to set the first derivitive equal to zero?

 

Indeed.