Some homework problems

[Monsta]

Hey guys, please help me with some of these questions, I'm really clueless how to solve them. I have a very large set of problems to to as homework and a big part I already solved but with these problems I don't have any idea. Any help is very much appreciated.

1: A ship S leaves a Port P at 2 pm and it sails in the direction 30°T at 12km/h. Another ship is located 100 km East of P and is sailing towards P at 8km/h. (t is the number of hours after 2 pm)

a: Show that the distance D(t), between the two ships is given by D(t) = Square root of: 304t²-2800t+10000

b: Find the minimum value of [D(t)]² for all t>0.

c: At what time are both ships closest?


2:A fence AB is 1m high and 2m away from a wall, RQ. An ladder PQ is placed on the ence and touches the ground at P and the wall at Q.

a: If AP = x, find the value of QR in terms of x.
b: The ladder has the length L(x), show that [L(x)]² = (x+2)²(1+(1/x²)

Show that derivative[L(x)]² with respect to x = 0, only when x = third square root of 2.

d: Find the shortest length of the ladder and prove that it is the shortest length.


3: A (symmetrical) gutter is made from a sheet of metal (30 cm wide) by bending it twice. For alpha as indicated:

a: Show that the cross-sectional area is given by A = 100cosalpha(1+cosalpha)
b: Using the result from a, show that derivative A with respect to alpha = 0 when sin alpha = 0.5 or -1.
c: What value has alpha when the gutter has its maximum carrying capacity?

Here I made a sketch for 2 of the problems:

 

[galactus]

1: A ship S leaves a Port P at 2 pm and it sails in the direction 30°T at 12km/h. Another ship is located 100 km East of P and is sailing towards P at 8km/h. (t is the number of hours after 2 pm)

a: Show that the distance D(t), between the two ships is given by D(t) = Square root of: 304t²-2800t+10000

b: Find the minimum value of [D(t)]² for all t>0.

c: At what time are both ships closest?

Use the law of cosines. \(\displaystyle \L\\D^{2}=a^{2}+b^{2}-2abcos(\frac{\pi}{3})\)

The second ship is initially 100 km away and getting closer at a rate of 8 km/h. Therefore, it's distance from P at any time is given by 100-8t.

Let \(\displaystyle \L\\a=100-8t, \;\ b=12t\)

You should get the quadratic D(t) they gave you.

Differentiate, set to 0 and solve for t. Then add that to 2 to find the time they are closest.

2:A fence AB is 1m high and 2m away from a wall, RQ. An ladder PQ is placed on the ence and touches the ground at P and the wall at Q.

a: If AP = x, find the value of QR in terms of x.

The idea is to use proportions. That is, similar triangles. Anytime you see a problem like this, use similar triangles.

The height of the wall is y. The distance from the base of the wall to where the ladder touches the ground is 2+x.

The fence is 1 meter high.

Now, we have all the info to use similar triangles. The big triangle formed
by PQR and the littler one formed by PBA.

\(\displaystyle \L\\\frac{1}{x}=\frac{y}{2+x}\)

Solve for \(\displaystyle \L\\y=\frac{x+2}{x}\)

b: The ladder has the length L(x), show that [L(x)]² = (x+2)²(1+(1/x²)

L(x) is just Pythagoras.

Show that derivative[L(x)]² with respect to x = 0, only when x = third square root of 2.

d: Find the shortest length of the ladder and prove that it is the shortest length.

Differentiate L^2 and set it to 0 and solve for x. Once you have that you can easily find the shortest ladder .


For the gutter problem, you have the area of a trapezoid formula.

\(\displaystyle \L\\A=\frac{1}{2}(10+b)h\)

Use basic trig (sin and cos) to express b and h as functions of alpha, then sub them into your trapezoid formula.

\(\displaystyle \L\\h=10cos{\alpha}, \;\ b=10+20sin{\alpha}\)

Sub them in the formula:

\(\displaystyle \L\\\frac{1}{2}(10+\underbrace{(10+20sin{\alpha})}_{\text{b}})(\underbrace{10cos{\alpha}}_{\text{h}})\)

See how I got those?.


Differentiate, set to 0 and solve for alpha. You should get 30 degrees or Pi/6

 

[Monsta]

I got it, thank you so much for you help!