1. ## 3 Calculus Problems

If x^2+y^2=8100 and dy/dt=3, find dx/dt when y=72.

A man starts walking north at 5ft/s from a point P. Five minutes later, a woman starts walking south at 4ft/s from a point 1300ft due east of P. At what rate are the people moving apart 20min after the woman starts walking?

A particle moves along the curve y=(19+x^3)^(1/2). As it reaches the point (5, 12), the y-coordinate is increasing at a rate of 3cm/s. How fast is the x-coordinate of the point changing at that instant?

Thank you!

2. Originally Posted by asnxbbyx113
If x^2+y^2=8100 and dy/dt=3, find dx/dt when y=72.
$y=72$ then $x=54$
$2x\frac{dx}{dt}+2y\frac{dy}{dx}=0$

Thus,
$2(54)\frac{dx}{dt}+2(72)(3)=0$
Solve.

3. #3:

$y=\sqrt{19+x^{3}}$

We want dx/dt when dy/dt=3.

$y^{2}-19=x^{3}$

Differentiate:

$3x^{2}\frac{dx}{dt}=2y\frac{dy}{dt}$

$\frac{dx}{dt}=\frac{2y}{3x^{2}}\frac{dy}{dt}$

$\frac{dx}{dt}=\frac{2(12)}{3(5)^{2}}(3)=\frac{24}{ 25}$

4. Originally Posted by asnxbbyx113

A man starts walking north at 5ft/s from a point P. Five minutes later, a woman starts walking south at 4ft/s from a point 1300ft due east of P. At what rate are the people moving apart 20min after the woman starts walking?

Thank you!
Umm, this is interesting.

We need a diagram or figure to understand the solution. You draw it on paper.

From a point P, draw a line segment vertically upwards up to a point B [for boy].
PB = distance the man walked in 5 minutes = (5 ft/sec)(5min *60sec/min) = 5*300 = 1500 ft.
Draw a line segment from point P to the right, perpendicular to PB, that is supposed to be 1300 ft long. Call its end, point G [for girl]. So PG is 1300 ft.

The girl/woman walks southward at 4ft/sec; the boy/man continues walking northwards at 5ft/sec.

Let x = distance the boy walks after point B at any time t sec.
And y = distance the girl walks from point G at the same time t sec.

Draw a line segment upwards from B up to, say, point A. So, BA = x.
Draw a line segment downwards from G down to, say, point C. So, GC = y.
Then draw a line connecting A and C. Line AC is the distance from the boy to the girl at any time t sec. Call AC as s. AC or s intersects PG at, say, point E.

Express AC or s in terms of x and y. Differentiate with respect to time t. The ds/dt is what we are asked to find.

I said in the beginning that this problem is interesting.
We have a figure where there are two right triangles---one, triangle APE, "points" upwards; the other, triangle CGE, "points" downward. Using this figure in expressing s in terms of x and y here, and differentiating later, one will need the patience and perseverance of Job [biblical person] or of a worker ant. We are no Job, nor worker ant, so we do not use this figure.

We modify the figure. There is an old trick in doing this. Out of the two right triangles, we make a single right triangle where the hypotenuse AC or s remains where it is.
Relocate y vertically below point P.
Relocate PG horizontally from the bottom of y.
So we have a new figure, a right triangle, with these:
---vertical leg = (x +1500 +y) ft.
---horizontal leg = 1300 ft.
---hypotenuse = s ft.

Very easy now.

By Pythagorean theorem,
s^2 = (x +1500 +y)^2 +(1300)^2 --------------(1)
Differentiate both sides with respect to time t,
2s(ds/dt) = 2(x +1500 +y)(dx/dt +dy/dt)
s(ds/dt) = (x +1500 +y)(dx/dt +dy/dt) ----------(2)

In (2) we know dx/dt = 5ft/sec and dy/dt = 4ft/sec. We use them as both positive, although they are opposite in directions, because s grows longer as x and y grow longer.
We don't know s, ds/dt, x and y.

We can find x, y and s dy using the distance formula distance = rate*time.

"At what rate are the people moving apart 20min after the woman starts walking."
So, t = 20 min = 20*60 = 1200 sec.
And so, x = 5*1200 = 6000 ft.
And, y = 4*1200 = 4800 ft
Substitute those in (1),
s^2 = (6000+1500+4800)^2 +(1300)^2
s = 12,368.5 ft.

Substitute those in (2),
s(ds/dt) = (x +1500 +y)(dx/dt +dy/dt) ----------(2)
(12,368.5)(ds/dt) = (6000+1500+4800)(5 +4)
ds/dt = (12,300)(9) / (12,368.5)
ds/dt = 8.95 ft/sec

Therefore, 20 minutes after the girl started walking, the boy and the girl are separating at the rate of 8.95 ft/sec. -----------answer.

===================
If you doubt the modification of the original two right triangles into a single right triangle, try playing with two 3-4-5 rigth triangles. One, a 6-8-10 triangle, the other, a 3-4-5 triangle. See for yourself.

5. Hello, asnxbbyx113!

Another approach to #2 . . .

2) A man starts walking north at 5ft/s from a point $P.$ .Five minutes later,
a woman starts walking south at 4ft/s from a point $Q$ 1300 ft due east of $P.$
At what rate are the people moving apart 20 min after the woman starts walking?

The man walks north from $P$ at 5 ft/sec.
Since he has a 5-minute (300-second) headstart.
. . he has already walked: $5 \times 300 \:=\:1500$ feet.
In the next $t$ seconds, he walks $5t$ feet more to $A.$

The woman walks south from $Q$ at 4 ft/s.
In the next $t$ seconds, she walks $4t$ feet to $B.$

The diagram looks like this:
Code:
      A *
| *
5t|   *
|     *
+       *
|         *  R
1500|           *
|             *
P *---------------*-------* Q
:                 *     |
4t:                   *   |4t
:                     * |
C + - - - - - - - - - - - * B
:         1300          :

The distance between them is: $R \,=\,AB.$

From the right triangle $ACB$, we have:
. . $R^2 \:=\:(5t + 1500 + 4t)^2 + 1300^2\quad\Rightarrow\quad R \:=\:\left[(9t + 1500)^2 + 1300^2\right]^{\frac{1}{2}}$

Differentiate: . $\frac{dR}{dt} \:=\:\frac{1}{2}\left[(9t + 1500)^2 + 1300^2\right]^{-\frac{1}{2}}\left[2(9t + 1500)\cdot9\right]$

. . and we have: . $\frac{dR}{dt} \:=\:\frac{9(9y + 1500)}{\sqrt{(9t + 1500)^2 + 1300^2}}$

"20 minutes after the woman starts walking" means: $t = 1200$ seconds.

. . Plug it in and simplify . . .