Need Help On Calculus

Damn I wish my school gave laptop, it would make life so much easier. My school tuition is just under $400/credit and then we got about another $400-$500 in "mandatory fees" we never make use of (except the bus pass).
I hate UH

 

control theory was one of my favorite courses, ...we had a lab portion to the class, ...our first lab project was determining the internal pressure of an unopened soda can using strategically placed strain gauges...

...you also get back to calc in any advanced vibrations class, i.e. structural dynamics, etc. ...waveform analysis like fourier transform, etc... also a course on optimization, finding minima and maxima of multidimensional cost functions...

 

I had to calculate the internal pressure of a soda can in Chemistry last semester. I think it was 3atm but I forget already.
Also calculated how long a room temperature beer takes to chill in a refrigerator in my pre-calculus class, I think it was 35 minutes.

 

Here's the next set. I have an exam on Friday so I'll be studying a lot.

For HW16, question 30(6) the one about the 6' guy walking to the streetlight:
I got the Rate of Change for his shadow is -6 ft/sec

For HW17, question 44(2) about the surveyor and the building:
I got the Maximum Allowed Error D(theta) = cos^2(75)/30

My instructor checked both and said they were correct. Any help on the others would be great.

 

HW16, number 28(6): I got (dy/dt)=(5root3)/3
I don't know how to answer number 36(5)

HW17, 4(1): I got -23/12(x+8)
6(2)a: x
6(2)b: x*sec^2(0)
I don't know number 36(1)

 

...i'm looking at 36(5)... i think i know the way... let me ask if you have been able to first formulate the equation for theta(t) ...i.e. theta as a function of time, ..hint: it's a function of x(t) and y(t), ..which are the equations for the two different movers...

 

That's where I'm stuck. I got dy/dt=-2 and dx/dt=1.
Did I at least get that much right?
I also used Pythagorean theorem to find the hypotenuse at 10root5 but Idk if I even need it.
I also figured with the distances of 10 and 20 meters, tangent theta is tan(1/2), but Idk how to use that to find the rate of change.

 

ok, as i see it (i.e., i don't know if this is the approach intended based on your current position in the class, etc...), but first you must forumlate an analytical form, equation, for theta.... so, theta(t) = f(x(t), y(t)), ...where x(t) is an equation for the person moving east-west, and y(t) is an equation for the person moving north-south... so, to distill this, what i'm saying is that theta is a function of x and y, which are both functions of time, thus theta is a function of time... this is a bit of a complex one...

so, here's my approach, see if you agree:

step 1: forumlate analytical form for x(t)...
step 2: forumulate analytical form for y(t)...
step 3: forumlate analytical form for theta(x(t),y(t))...
step 4: compute analytical derivative of theta(t), i.e. theta-dot(t) "dot" means time-derivative...

step 5: plug-n-chug the provided values into the theta-dot(t) equation...

follow me?

...see if you can get step 1 and 2, first...

 

Darn I turned it in right before u posted this. I'll get my paper back tomorrow and see how it went.
I think the answer I got was something like -2/sec^2(1/2) rad/sec.
It was a desperation attempt haha

 

Two more I'm doing today, they're due tomorrow.

Anybody got tips to help me out?

 

So far I have...

HW24
2(7): Height and Width are 2in.

4(7): Height is 8in., Width is 4in.

8(7): Length of fence is Square root 108 meters


HW25
14(7): Height is 20/cube root pi, Radius is 10/cube root pi

32(7): I'm still working on

34(7): Height is root 96, Width is root 48