Ask Me Help Desk - Distance and speed

Please help me to solve the problem. Thanks. Please show me the full solution. Appreciate it a lot. Hereby I attached the question.

Quote:

Originally Posted by JiunLing

Please help me to solve the problem. Thanks. Please show me the full solution. Appreciate it a lot. Hereby I attached the question.

Interesting question. I would also like to know the answer. At first I thought that it was a Pythagorean question. When the rocket is at 300 meters you would be 500 meters from it at that instant. I guess it all depends on what you mean by, 'at that instant'.

Alternatively you would need to come up with an equation involving distance height over distance time.

Tut

Tut: it is a pythagorean question, it just also uses calculus.

JuinLing: Okay, first you need the distance as a function of the height of the rocket. That's not to hard, it just uses the pythagorean theorem. Since the derivative is just the instantaneous change, then we're going to take the derivative of the distance (change in distance) in terms of the change in height (which is just the velocity). Then we just plug all our numbers in, and then we'll have the answer.

Let $D$ be the distance rom the rocket and let $H$ be the height. So the distance in terms of the height is:
$D = \sqrt{H^2 +400^2}$

Let $\Delta D$ be the change in distance, and let $\Delta H$ be the change in height. By applying the chain rule to our distance function we get:

$\Delta D = \frac{2H \cdot \Delta H }{2 \cdot \sqrt{H^2 +400^2} \cdot$

$\, = \frac{H\cdot \Delta H}{\sqrt{H^2 +400^2}$ .

Now all we need to do is plug in $H$ and $\Delta H$ and we're done. So we get:

$\Delta D = \frac{300 \cdot 600}{\sqrt{300^2 +400^2}$

$\, = \frac{180000}{\sqrt{90000 +160000}$

$\,= \frac{180000}{\sqrt{250000}$

$\,= \frac{180000}{500}$

$\, = 360$

So our answer is the distance from the rocket is changing at a rate of 360m/s at the instant the rocket is at 300m.

I don't know what level of math you're at, so if my use of the chain rule is a little confusing, please let me know, and I'll do my best to explain it.

I hope this helps :)

Quote:

Originally Posted by corrigan

Tut: it is a pythagorean question, it just also uses calculus.

JuinLing: Okay, first you need the distance as a function of the height of the rocket. That's not to hard, it just uses the pythagorean theorem. Since the derivative is just the instantaneous change, then we're going to take the derivative of the distance (change in distance) in terms of the change in height (which is just the velocity). Then we just plug all our numbers in, and then we'll have the answer.

Let $D$ be the distance rom the rocket and let $H$ be the height. So the distance in terms of the height is:
$D = \sqrt{H^2 +400^2}$

Let $\Delta D$ be the change in distance, and let $\Delta H$ be the change in height. By applying the chain rule to our distance function we get:

$\Delta D = \frac{2H \cdot \Delta H }{2 \cdot \sqrt{H^2 +400^2} \cdot$

$\, = \frac{H\cdot \Delta H}{\sqrt{H^2 +400^2}$ .

Now all we need to do is plug in $H$ and $\Delta H$ and we're done. So we get:

$\Delta D = \frac{300 \cdot 600}{\sqrt{300^2 +400^2}$

$\, = \frac{180000}{\sqrt{90000 +160000}$

$\,= \frac{180000}{\sqrt{250000}$

$\,= \frac{180000}{500}$

$\, = 360$

So our answer is the distance from the rocket is changing at a rate of 360m/s at the instant the rocket is at 300m.

I don't know what level of math you're at, so if my use of the chain rule is a little confusing, please let me know, and I'll do my best to explain it.

I hope this helps :)

Hi corrigan,

I used H= the square root of d squared + 400 squared

d=600t

= square root of 600 squared t squared +400 squared.

Basically I found the square root of that and got 360 meters per second.

Thanks

Tut

How to get the formula of the change of D?
https://www.askmehelpdesk.com/cgi-bin/mimetex.cgi?%20\Delta%20D%20=%20\frac{2H%20\cdot%2 0\Delta%20H%20}{2%20\cdot%20\sqrt{H^2%20+400^2}%20 \cdot

I got the change in $D$ by taking the derivative of $D$ . Remember, $D$ was simply a function in terms of the height. Perhaps a less confusing way to look at it would be:

$D = \sqrt{H^2 +400^2}$ (think of $D$ as $y$ , and $H$ as $x$ )

$\Rightarrow \frac{\Delta D}{\Delta H } = \frac{H}{\sqrt{H^2 +400^2}$

$\Rightarrow \Delta D= \frac{H \Delta H }{\sqrt{H^2 +400^2}$

All a derivative is is the change in the function relative to the change in the argument (an argument in a function is usually a variable). For example, take the function $y = 3x +2$ , then $\frac{dy}{dx} = 3$ . All this means is that as $x$ changes, $y$ changes three times as much. Sometimes the change (or derivative) is a function in terms of the variable. Take for example $y=x^2 +3x+2$ , then $\frac{dy}{dx} = 2x+3$ . So $\frac{dy}{dx}$ is a function in terms of $x$ , what that means is that the amount that $y$ changes depends not only on the change in $x$ , but also on the value of $x$ at whatever point we are looking at. It's the nuts and bolts of the chain rule that a lot of calculus teachers don't tell there students for a variety of reasons.

I'll be keeping an eye on this thread. If you need more help, just post it with this question.

Tut: Sorry, I didn't see your answer before my last post. The answer was right, but the way you got there was imprecise. Your solution implies that the rocket was going at a constant velocity of 600 m/s. When dealing with trajectories there is a lot of different factors, acceleration, drag, initial velocity, etc. all the question gave was the current height and velocity. I don't know JuinLing's level of calculus, so I was trying to give an answer that was as dynamic as possible to help him answer similar questions.

Quote:

Originally Posted by corrigan

Tut: Sorry, I didn't see your answer before my last post. The answer was right, but the way you got there was imprecise. Your solution implies that the rocket was going at a constant velocity of 600 m/s. When dealing with trajectories there is a lot of different factors, acceleration, drag, initial velocity, etc. all the question gave was the current height and velocity. I don't know JuinLing's level of calculus, so I was trying to give an answer that was as dynamic as possible to help him answer similar questions.

Hi corrigan,

Yes, that is what I assumed. I wasn't really sure about that bit. Thanks for the clarification.

TUt

Thanks... I get it already... just differentaite the equation... haha^^ I'm blur with my examination.