calculus problem

**asian5** · July 31st 2006, 11:38 PM

Hi evry1, this is a problem i encountered and i cant seem to solve it. can you please help me out. I would really appreciate it if you could help me with all the questions, as im really struggling!!! thanks (sorry about the vertical lines, they are meant to be solid, but i cudnt make solid lines on the computer, and ignore the full stops, as they are included becos you cant just use spacing)

..........start boundary
S _____________________
|.angle Y........sand.........|
|..................................| 200meters
|------g------- X -------|
|..mud..........................| 200meters
|______________angle Z_|
........^ 500meters..........F
........^
........^
........^
........^
.....finish boundary

There is a race that goes through sand and mud, from S to F, passing through X as shown. Kylie can run at 4.5meters/second through sand and 2.5meters/second through mud. The track is the hypotenues from S to X, then the hypotenus from X to F. There is an angle Y, which is between the track (hypotenus from S to X) and the start boundary and also angle Z, which is between the track (hypotenus from X to F) and the finish boundary.

Question 1)
show that the total distance covered in the race, d=200(secY+secZ) and prove that Y and Z are also related by the equation tanY+tanZ=K, where 'K' is a constant.

Question 2)
Find an expression for d in terms of Y and state the implied domain of the corresponding function. Find the total distance kylie runs if she starts off at an angle of 50degrees(Y) and state the angleZ she must use when she reaches the second surface

Question 3)
Find an expression for t, the time taken for her to travel from S to F, in terms of Y and state the implied domain of the corresponding function. Plot the graph of t against g (the distance from the S vertical boundary and X) and comment on what the graph shows.

Question 4)
Another contestant, Emily, can run 4meters/second through sand and 3meters/second through mud. Determine Emily's best route, ensuring that she starts at point S and finishes at point F, passing through a point X, but it doesnt necessarily have to be the same point that Kylie passed through. Who wins the contest and by how many seconds?

Question 5)
Now let t1= the time taken to travel from S to X along the hypotenuse
and let t2= the time taken to travel from X to F along the hypotenuse
and t= t1 + t2

Express t1 and t2 in terms of 'g' and find dt1/dg (derivative) and dt2/dg (derivative) in terms of g

Question 6)
With the angles of Y and Z as shown originally in the figure, express dt1/dg and dt2/dg in terms of Y and Z respectively

Question 7)
Hence express dt/dg in terms of Y and Z and show that the minimum time occurs when (sinY/a) = (sinZ/b) where a and b are constants. Give an interpretation of these constants in terms of the situation being modelled.

Question 8)
Solve (sinY/a) = (sinZ/b) and tanY+tanZ=K to find the values of Y and Z which simultaneously satisfy each of these equations.

Question 9)
Hence verify that the best route for Kylie is the same as that found previously

**Quick** · August 1st 2006, 04:48 AM

Is this what your diagram means:

$S$
$\underbrace{\boxed{\begin{array}{cccccc}Y&#&Sand&# \\#\\#&#&X---&-- \\#&Mud&#\\#&#&#&Z\end{array}}}_{500m}$ $\begin{array}{cccccccc}#\\#\\#\\200m\\#\\#\\200m\\ F<br />$

**Soroban** · August 1st 2006, 02:10 PM

Hello, asian5!

Are there typos in the problem?
. . I can't get past Question #1 . . .

I believe your diagram looks like this:

Code:

    S o-----------------------+
      |   * Y°         (sand) |
  200 |       *               |
      |           *   X       |
    A + - - - - - - - o - - - + B
      |                 *     |
  200 |                   *   |
      | (mud)            Z° * |
      +-----------------------o F
      : - - - -  500m - - - - :

This race goes through sand and mud, from $S$ to $F$ , passing through $X$ as shown.
Kylie can run at 4.5 m/sec through sand and 2.5 m/sec through mud.
The track is the hypotenuse from $S$ to $X$ , then the hypotenuse from $X$ to $F.$
There is an angle $Y$ , which is between the track $SX$ and the start boundary
and angle $Z$ , which is between the track $XF$ and the finish boundary.

Q1) Show that the total distance covered in the race is: $d \:=\:200(\sec Y+\sec Z)$ ?
and prove that Y and Z are also related by the equation: $\tan Y+\tan Z\:=\:K$ ?
where $K$ is a constant.

If my diagram is correct (and angles $Y$ and $Z$ are correctly placed),
. . then these two statements are not true.

We have: . $\begin{Bmatrix}\csc Y = \frac{SX}{200}\quad\Rightarrow\quad SX = 200\csc Y \\ \\ \csc Z = \frac{XF}{200}\quad\Rightarrow\quad XF = 200\csc Z\end{Bmatrix}$

Add: . $SX + XF \:=\:200\csc Y + 200\csc Z$

. . Therefore: . $\boxed{d \;= \; 200(\csc Y + \csc Z)}$

We have: . $\begin{Bmatrix}\cot Y = \frac{AX}{200}\quad\Rightarrow\quad 200\cot Y = AX \\ \\ \cot Z = \frac{XB}{200}\quad\Rightarrow\quad 200\cot Z = XB\end{Bmatrix}$

Add: . $200\cot Y + 200\cot Z\:=\:AX + XB$

. . Therefore: . $200\cot Y + 200\cos Z\:=\:500\quad\Rightarrow\quad \boxed{\cot Y + \cot Z = \frac{5}{2}}$

[So I don't dare try the other parts . . . ]

**asian5** · August 2nd 2006, 12:03 AM

its almost correct, thanks for your help. The length 'a' which was stated originally is between you diagrams A and X. And the angles Y and Z are actually 90degress-your angle, so on the opposite side of the hypotenuse. I wud really appreciate your continued help. thanks very much

**Quick** · August 2nd 2006, 04:19 AM

This race goes through sand and mud, from S to F, passing through X as shown.
Kylie can run at 4.5 m/sec through sand and 2.5 m/sec through mud.
The track is the hypotenuse from S to X, then the hypotenuse from X to F.
There is an angle Y, which is between the track SX and the start boundary
and angle Z, which is between the track XF and the finish boundary.

Q1) Show that the total distance covered in the race is: d \:=\:200(\sec Y+\sec Z)
and prove that Y and Z are also related by the equation: \tan Y+\tan Z\:=\:K
where K is a constant.

Alright, I have never learned "sec" before, but I'm going to assume that it means $\frac{hypotenuse}{adjacent}$

if that assumption is correct than the rest is easy, to find the hypotenuse using sec you'll get $adjacecnt\times\sec A$

therefore the distance from S to X is $\sec Y=\frac{hypotenuse}{adjacent}\quad\Rightarrow\quad \sec Y=$ $\frac{\overline{SX}}{200}\quad\Rightarrow\quad 200\sec Y=\overline{SX}$

Now we find the distance from X to F
$\sec Z=\frac{hypotenuse}{adjacent}\quad\Rightarrow\quad \sec Z=$ $\frac{\overline{XF}}{200}\quad\Rightarrow\quad 200\sec Z=\overline{XF}$

now we combine those two to get the total distance:
$d=200\sec Y+200\sec Z \quad\Rightarrow\quad d=200(\sec Y + \sec Z)$

alright, now we need to prove that $\tan Y +\tan Z=K$ to do that we need to find what tan is:
$\tan=\frac{opposite}{adjacent}$
so therefore: $adjacent \times\tan=opposite$

so if we wrote $200\tan Y$ we would find length a, and if we wrote $200\tan Z$ we would find length b. We know length a and b together equal 500, so we write it out like this:

$a+b=500$

substitute: $200\tan Y+200\tan Z=500$

divide by 200: $\tan Y+\tan Z=\frac{5}{2}$

**topsquark** · August 2nd 2006, 04:55 AM

For the record:

$csc( \theta ) = 1/sin( \theta )$ (csc = "cosecant")
$sec( \theta ) = 1/cos( \theta )$ (sec = "secant")
$cot( \theta ) = 1/tan( \theta )$ (cot = "cotangent")

-Dan

**Soroban** · August 2nd 2006, 05:13 AM

Hello, asian5!

Okay, I think I've got the diagram now.
But the problems are long and intricate.
. . I can do only a few at a time . . .

Code:

    S o-----------------------+
      |Y° *                   |
  200 |       *               | 200
      |           *     500-a |
    A + - - - - - - - o - - - + B
      |       a       X *     |
  200 |                   * Z°| 200
      |                     * |
      +-----------------------o F
      : - - - -  500m - - - - :

1)a) Show that the total distance covered in the race is: $d \:=\:200(\sec Y + \sec Z)$
b) Prove that $Y$ and $Z$ are also related by the equation: $\tan Y + \tan Z\:=\:K$
where $K$ is a constant.

We have: . $\sec Y = \frac{SX}{200}\quad\Rightarrow\quad SX = 200\sec Y\;\;\bf{[1]}$

. . . .and: . $\sec Z = \frac{XF}{200}\quad\Rightarrow\quad XF = 200\sec Z\;\;\bf{[2]}$

Add [1] and [2]: . $SX + XF\:=\:200\sec Y + 200\secZ$

. . Therefore: . $1(a)\;\;\boxed{d \;= \; 200(\sec Y + \sec Z)}$

We have: . $\begin{array}{ccc}\tan Y \,= \,\frac{a}{200} \\ \\ \tan Z \,= \,\frac{500-a}{200}\end{array}\;\begin{array}{ccc}[3] \\ \\ [4]\end{array}$

Add [3] and [4]: . $\tan Y + \tan Z\:=\:\frac{a}{200} + \frac{500-a}{200} \:=\:\frac{500}{200}$

. . Therefore: . $1(b)\;\;\boxed{\tan Y + \tan Z \;=\;\frac{5}{2}}$

2)(a) Find an expression for d in terms of Y and state the implied domain of the function.
(b) Find the total distance Kylie runs if she starts off at an angle of $y = 50^o$
(c) State the angle $Z$ she must use when she reaches the second surface.

Part (a) is quite ugly . . . maybe someone can find a neater approach.

We have: . $\tan Y + \tan Z \:=\:\frac{5}{2}\quad\Rightarrow\quad\tan Z\:=\:\frac{5}{2} - \tan Y$

Then: . $\sec^2Z \:=\:1 + \tan^2Z\:=\:1 + \left(\frac{5}{2} - \tan Y\right)^2$

which simplifies to: . $\sec Z \:=\:\frac{\sqrt{4\sec^2Y - 20\tan Y + 25}}{2}$

. . Therefore: . $2(a)\;\;\boxed{d \;= \;200\left[\sec Y + \frac{1}{2}\sqrt{4\sec^2Y - 20\tan Y + 25}\right]}$

The domain is rather obvious but: . $\boxed{0^o\,<\,Y\,<\,90^o}$

If $Y = 50^o$ , then: . $2(b)\;\;d\;=\;200\left[\sec50^o + \frac{1}{2}\sqrt{4\sec^250^o - 20\tan50^o + 25}\right]$

I got $640.5$ feet . . . but please check my work . . . all my work.

If $Y = 50^o$ , we have: . $\tan50^o + \tan Z\:=\:\frac{5}{2}$

then: . $\tan Z\:=\:\frac{5}{2} - \tan50^o\:=\;1.308246407$

Hence: . $\;Z \:=\:\tan^{-1}(1.308246407) \:=\:52.00635495$

. . Therefore: . $2(c)\;\; \boxed{Z\:\approx\:52^o}$

**asian5** · August 3rd 2006, 02:50 PM

thanks this is looking great!!!!
could you please continue...
many thanks

**Quick** · August 3rd 2006, 04:26 PM

Originally Posted by Soroban

Part (a) is quite ugly . . . maybe someone can find a neater approach.

maybe this way's better:

remember: $\sec\theta=\frac{hypotenuse}{adjacent}$

therefore: $\sec Z=\frac{\sqrt{200^2+(500-a)^2}}{200}$

therefore: $\sec Z=\frac{\sqrt{40000+250000-1000a+a^2)}}{200}$

then: $\sec Z=\frac{\sqrt{40000+250000-1000a+a^2)}}{200}$

substitute 200tanY for a: $\sec Z=\frac{\sqrt{290000-1000(200\tan Y)+(200\tan Y)^2}}{200}$

multiply: $\sec Z=\frac{\sqrt{210000-200000\tan Y+40000\tan^2 Y}}{200}$

therefore: $\sec Z=\sqrt{\frac{210000-200000\tan Y+40000\tan^2 Y}{40000}}$

divide: $\sec Z=\sqrt{5.25-5\tan Y+\tan^2 Y}$

factor: $\sec Z=\sqrt{5.25-\tan Y(5-\tan Y)}$

then substitute into distance: $\boxed{d=200(\sec Y+\sqrt{5.25-\tan Y(5-\tan Y)}}$

or maybe this way's just as bad...

but the answer for y=50 comes out as 479.867

who do you think's right?

**ThePerfectHacker** · August 3rd 2006, 07:11 PM

asian stop begging for answers!

You can wait!!!

-=USER WARNED=-
(Message soon to be deleted)

**asian5** · August 3rd 2006, 07:12 PM

i think the distance should be greater than 620 when y=50, becos 620 is the direct hypotenuse from S to F, which is the shortest distance. Therefore i think soroban is correct with his 640. but i still apreciate evry1s help. please continue

**Quick** · August 3rd 2006, 07:51 PM

Originally Posted by asian5

Question 3)
Find an expression for t, the time taken for her to travel from S to F, in terms of Y and state the implied domain of the corresponding function. Plot the graph of t against g (the distance from the S vertical boundary and X) and comment on what the graph shows.

for this one we need to make a new equation

the question asks us what her speed is compared to the line segment a

this equation though is surprisingly easy: we need to find the hypotenuse of triangle AXY, which is $\sqrt{200^2+a^2}=\sqrt{40000+a^2}$

now we need to find her speed in terms of seconds (remember when she's running on this hypotenuse she's running on sand), which is: $\frac{\sqrt{40000+a^2}}{4.5}$

now repeat the process for 500-a and you'll get:

$\sqrt{200^2+(500-a)^2}$ = $\sqrt{40000+250000-1000a+a^2}=\sqrt{290000-1000a+a^2}$

and add speed (this one's through mud)

$\frac{\sqrt{40000+250000-1000a+a^2}}{2.5}$

now add the two together: $t=\frac{\sqrt{40000+a^2}}{4.5}+\frac{\sqrt{290000-1000a+a^2}}{2.5}$

strangely enough, they call "a" "g" in this problem so write it all out:

$t=\frac{\sqrt{40000+g^2}}{4.5}+\frac{\sqrt{290000-1000g+g^2}}{2.5}$

and graph (I'll let you comment):
note: the legend should say total time

PS the remaining questions actually require calculus so I can't help you with them

**Quick** · August 3rd 2006, 08:19 PM

Originally Posted by asian5

i think the distance should be greater than 620 when y=50, becos 620 is the direct hypotenuse from S to F, which is the shortest distance. Therefore i think soroban is correct with his 640. but i still apreciate evry1s help. please continue

This seems logical, but I can't figure out what's wrong with my work!

**asian5** · August 3rd 2006, 10:35 PM

g is meant to be a. sorry for the mistake in typing. But it says to get t in terms of the angle Y.

**Quick** · August 4th 2006, 06:06 AM

Originally Posted by asian5

g is meant to be a. sorry for the mistake in typing. But it says to get t in terms of the angle Y.

my equation does call "a" "g"

if you want it in terms of Y than just substituti 200tanY for g

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