# Thread: Calc I word problems help

1. 1) A space shuttle launches with altitude function a(t) = 20t^2 (t = seconds and a(t) = meters). An observer standing 4 miles away (horizontally) from the launch pad must look up at a higher and higher angle as time goes by in order to watch the space shuttle. Calculate the rate of change of that angle after 1 second of lift-off.

2) Suppose that a traffic drone is 1000 feet in the air, directly over a straight freeway.
It observes a car on the freeway that is 2000 feet away from the drone traveling
away from the drone at 80 feet per second. The speed limit is 65 miles per hour.

3) A coffee filter has the shape of an inverted cone. Water drains out of the filter at
a rate of 10cm^3 per minute. When the depth of the water in the cone is 8cm, the
depth is decreasing at a rate of 2cm per minute. What is the ratio of the height of  Reply With Quote

2.

3. Hello and welcome to MHB, fidal2! We ask that our users show their progress (work thus far or thoughts on how to begin) when posting questions. This way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Just for future reference, we also ask that no more than 2 questions be asked per thread in the initial post. This way a thread does not potentially become convoluted and hard to follow.

Can you post what you have done so far?  Reply With Quote

4. Hello, fidal2! Quote:
2) Suppose that a traffic drone is 1000 feet in the air, directly over a straight freeway.
It observes a car on the freeway that is 2000 feet away from the drone
traveling away from the drone at 80 feet per second.
The speed limit is 65 miles per hour.

The drone is at $A:\: AB = 1000$

The car is at $C:\:x = BC.$

Let $y\,=\,AC.$

We have: $\:x^2 + 1000^2\:=\:y^2$

Differentiate with respect to time:
$\quad 2x\dfrac{dx}{dt}\:=\:2y\dfrac{dy}{dt} \quad\Rightarrow\quad \dfrac{dx}{dt} \:=\:\dfrac{y}{x}\,\dfrac{dy}{dt}$

We are given: $\:\frac{dy}{dt} = 80$ ft/sec.
When $y = 2000,\;x = 1000\sqrt{3}.$

Hence: $\:\dfrac{dx}{dt} =\:\dfrac{2000}{1000\sqrt{3}}\cdot 80 \;\approx\;92.376\text{ ft/sec} \;\approx\;63\text{ mph}$

No, the car is not breaking the speed limit.  Reply With Quote

5. Since several days has gone by with no feedback from the OP, I will post solutions to the remaining problems for the benefit of future readers. Originally Posted by fidal2 1) A space shuttle launches with altitude function a(t) = 20t^2 (t = seconds and a(t) = meters). An observer standing 4 miles away (horizontally) from the launch pad must look up at a higher and higher angle as time goes by in order to watch the space shuttle. Calculate the rate of change of that angle after 1 second of lift-off.
If we let $\theta$ be the angle of inclination, then we may state:

$\displaystyle \tan(\theta)=\frac{a(t)\text{ m}}{4\text{ mi}}\cdot\frac{2.54\text{ in}}{1\text{ cm}}\cdot\frac{1 \text{ft}}{12\text{ in}}\cdot\frac{1\text{mi}}{5280\text{ ft}}\cdot\frac{100\text{ cm}}{1\text{ m}}=\frac{127}{6336}t^2$

Differentiating with respect to time $t$, we obtain:

$\displaystyle \sec^2(\theta)\d{\theta}{t}=\frac{127}{3168}t$

Using a Pythagorean identity, we may write:

$\displaystyle \sec^2(\theta)=\tan^2(\theta)+1=\left(\frac{127}{6336}t^2\right)^2+1$

And so we obtain:

$\displaystyle \d{\theta}{t}=\frac{\dfrac{127}{3168}t}{\left(\dfrac{127}{6336}t^2\right)^2+1}=\frac{1609344t}{16129t^4+40144896}$

Thus:

$\displaystyle \left.\d{\theta}{t}\right|_{t=1}=\frac{1609344}{40161025}\,\frac{\text{rad}}{\text{s}}$ Originally Posted by fidal2 3) A coffee filter has the shape of an inverted cone. Water drains out of the filter at
a rate of 10cm^3 per minute. When the depth of the water in the cone is 8cm, the
depth is decreasing at a rate of 2cm per minute. What is the ratio of the height of
Let's let $k$ be the ratio of the height $h$ of the cone to its radius $r$, thus:

$\displaystyle r=\frac{h}{k}$

And so the volume of the cone is:

$\displaystyle V=\frac{1}{3}\pi h\left(\frac{h}{k}\right)^2=\frac{\pi}{3k^2}h^3$

Differentiating with respect to time $t$, there results:

$\displaystyle \d{V}{t}=\frac{\pi}{k^2}h^2\d{h}{t}$

Solving for $k$, we find:

$\displaystyle k=h\sqrt{\frac{\pi\d{h}{t}}{\d{V}{t}}}$

Plugging in the given data:

$\displaystyle h=8\text{ cm},\,\d{h}{t}=-2\,\frac{\text{cm}}{\text{min}},\,\d{V}{t}=-10\,\frac{\text{cm}^3}{\text{min}}$

We then find:

$\displaystyle k=8\sqrt{\frac{\pi}{5}}$  Reply With Quote

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