1. ## differentiation question

Hi, I'm new to this forum. Any help would be greatly appreciated. I'm actually tutoring a refugee in a volunteer program, but it's been so long since I've done this maths that I'm having real trouble with it so in order to help her, I myself need some help!

A Queenland resort has a large swimming pool with AB=75. The pool is rectangular and has corners A,B,C,D, with A and C as opposites and B and D as opposites. P is a point somewhere between D and C.
A boy can swim at 1 m/s and run at 1, 2/3 m/s. He starts at A, swims to a point P on DC, and runs from P to C. He takes 2 seconds to pull himself out of the pool.
a, Let DP=x m and the total time be T s. Show that T=square root of (x squared +900)+3/5 (75-x)+2.
b, find dT/dx
c. i Find the value of x for which the time taken is a minimum.
ii Find the minimum time.
d. Find the time taken if the boy runs from A to D and then D to C.

----------

An isosceles trapezoid is inscribed in the parabolas y=4-x squared as illustrated (the parabola intersects the x axis at (-2,0) and (2,0) and is symmetrical.
a. show that the area of the trapezoid is :
1/2(4-x squared) (2x+4)
b, show that the trapezoid has its greatest area when x=2/3
c, Repeat with the parabolas y = a squared - x squared
i, show that the area, A, of the trapezoid = ( a squared - x squared) ( a + x)
ii, use the product rule to find dA/ dx.
iii, show that a maximum occurs when x = a/3.

Thanks so much.

2. Originally Posted by afan17
...

An isosceles trapezoid is inscribed in the parabolas y=4-x squared as illustrated (the parabola intersects the x axis at (-2,0) and (2,0) and is symmetrical.
a. show that the area of the trapezoid is :
1/2(4-x squared) (2x+4)
b, show that the trapezoid has its greatest area when x=2/3
c, Repeat with the parabolas y = a squared - x squared
i, show that the area, A, of the trapezoid = ( a squared - x squared) ( a + x)
ii, use the product rule to find dA/ dx.
iii, show that a maximum occurs when x = a/3.
Let a and b denote the parallel sides of the trapezium and h it's height. Then the area of the trapezium is calculated by:

$A_t=\frac{a+b}2 \cdot h$

With your trapezium the base parallel has the length 4 and the upper parallel has the length 2x. The height correspond to the y-value. Plug in these terms into the formula given above:

$A_T(x)=\frac{2x+4}2 \cdot (4-x^2)$

which is exactly the given result.

To get the maximum value of $A_T$ calculate the derivative with respect to x:

$A_T(x)= (x+2)(4-x^2)=-x^3-2x^2+4x+8$

$A'_T(x)=-3x^2-4x+4$

$A'_T(x)=0~\implies~x=\frac23~\vee~x=-2$

The negative solution isn't very plausible here.

Do the following questions in exactly the same manner.

3. Hello, afan17!

I'm new to this forum . . . Welcome aboard!
I'm actually tutoring a refugee in a volunteer program . . . Good for you!

A resort has a rectangular swimming pool ABCD with AB = 75m, AD = 30m.
P is a point somewhere between D and C.
A boy can swim at 1 m/s and run at 1 2/3 m/s.
He starts at A, swims to point P, and runs from P to C.
He takes 2 seconds to pull himself out of the pool.

a) Let DP = x meters and the total time be T seconds.
Show that: . $T\:=\:\sqrt{x^2+900}+ \frac{3}{5}(75-x)+2$
Code:
      A             75              B
* - - - - - - - - - - - - - - *
|  *                          |
|     *                       |
30 |        *                    |
|           *                 |
|              *              |
* - - - - - - - - * - - - - - *
D        x        P   75-x   C
He will swim the diagonal distance $AP.$
This is the hypotenuse of right triangle $ADP.$
. . Hence:. . $AP \:=\:\sqrt{x^2 + 30^2}$
At 1 m/sec, this takes him: . $\frac{\sqrt{x^2+900}}{1} \:= \:{\color{blue}\sqrt{x^2+900}}$ seconds.

He runs the distance $PC \:=\:75-x\,\text{ at }\,\frac{5}{3}$ m/sec.
This takes him: . $\frac{75-x}{\frac{5}{3}} \:=\:{\color{blue}\frac{3}{5}(75-x)}$ seconds.

He also used 2 seconds to leave the pool.

His total time is: . $\boxed{T \;=\;\sqrt{x^2+900} + \frac{3}{5}(75-x) + 2}$ seconds.

b) Find $\frac{dT}{dx}$
We have: . $T \;=\;\left(x^2+900\right)^{\frac{1}{2}} - \frac{3}{5}x + 47$

Then: . $\frac{dT}{dx} \;=\;\frac{1}{2}\left(x^2+900\right)^{-\frac{1}{2}}(2x) - \frac{3}{5} \;=\;\boxed{\frac{x}{\sqrt{x^2+900}} - \frac{3}{5}}$

c) (i) Find the value of $x$ for which $T$ is a minimum.
(ii) Find the minimum time.
(i) Solve $\frac{dT}{dx} \:=\:0$

We have: . $\frac{x}{\sqrt{x^2+900}} \:=\:\frac{3}{5} \quad\Rightarrow\quad 5x\:=\:3\sqrt{x^2+900}$

Square both sides: . $25x^2 \:=\:9(x^2 + 900)\quad\Rightarrow\quad 25x^2 \:=\:9x^2 + 8100$

. . $16x^2 \:=\:8100\quad\Rightarrow\quad x^2 \:=\:\frac{2025}{4}\quad\Rightarrow\quad x \:=\:\frac{45}{2} \:=\:\boxed{22.5\text{ meters}}$

(ii) Substitute into the formula in part (a).

$T \;=\;\sqrt{22.5^2 + 900} + \frac{3}{5}(75-22.5) + 2 \;\;=\;\;\boxed{71\text{ seconds}}$

d) Find the time taken if the boy runs from A to D and then D to C.
He would run: . $30 + 75$ meters at $\frac{5}{3}$ m/sec

This will take him: . $\frac{105}{\frac{5}{3}} \:=\:\boxed{63\text{ seconds}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Very funny!

If the boy's only concern is getting from $A$ to $C$ as fast as possible,
. . he should all the way (and skip the math).

However, if this some sport where some swimming is required,
. . then our solution provides the shortest time.

4. Originally Posted by Soroban
Hello, afan17!

Code:
      A             75              B
* - - - - - - - - - - - - - - *
|  *                          |
|     *                       |
30 |        *                    |
|           *                 |
|              *              |
* - - - - - - - - * - - - - - *
D        x        P   75-x   C
He will swim the diagonal distance $AP.$
This is the hypotenuse of right triangle $ADP.$
. . Hence:. . $AP \:=\:\sqrt{x^2 + 30^2}$
At 1 m/sec, this takes him: . $\frac{\sqrt{x^2+900}}{1} \:= \:{\color{blue}\sqrt{x^2+900}}$ seconds.

He runs the distance $PC \:=\:75-x\,\text{ at }\,\frac{5}{3}$ m/sec.
This takes him: . $\frac{75-x}{\frac{5}{3}} \:=\:{\color{blue}\frac{3}{5}(75-x)}$ seconds.

He also used 2 seconds to leave the pool.

His total time is: . $\boxed{T \;=\;\sqrt{x^2+900} + \frac{3}{5}(75-x) + 2}$ seconds.

We have: . $T \;=\;\left(x^2+900\right)^{\frac{1}{2}} - \frac{3}{5}x + 47$

Then: . $\frac{dT}{dx} \;=\;\frac{1}{2}\left(x^2+900\right)^{-\frac{1}{2}}(2x) - \frac{3}{5} \;=\;\boxed{\frac{x}{\sqrt{x^2+900}} - \frac{3}{5}}$

(i) Solve $\frac{dT}{dx} \:=\:0$

We have: . $\frac{x}{\sqrt{x^2+900}} \:=\:\frac{3}{5} \quad\Rightarrow\quad 5x\:=\:3\sqrt{x^2+900}$

Square both sides: . $25x^2 \:=\:9(x^2 + 900)\quad\Rightarrow\quad 25x^2 \:=\:9x^2 + 8100$

. . $16x^2 \:=\:8100\quad\Rightarrow\quad x^2 \:=\:\frac{2025}{4}\quad\Rightarrow\quad x \:=\:\frac{45}{2} \:=\:\boxed{22.5\text{ meters}}$

(ii) Substitute into the formula in part (a).

$T \;=\;\sqrt{22.5^2 + 900} + \frac{3}{5}(75-22.5) + 2 \;\;=\;\;\boxed{71\text{ seconds}}$

He would run: . $30 + 75$ meters at $\frac{5}{3}$ m/sec

This will take him: . $\frac{105}{\frac{5}{3}} \:=\:\boxed{63\text{ seconds}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Very funny!

If the boy's only concern is getting from $A$ to $C$ as fast as possible,
. . he should all the way (and skip the math).

However, if this some sport where some swimming is required,
. . then our solution provides the shortest time.
Thank you so much! Honestly, such a lifesaver and it really is appreciated. Now I can go through this with the student tomorrow