# Thread: Analytical geometry

1. ## Analytical geometry

I tried these problems out but I am having too much trouble

A rock dropped out into a pond sends out a circular ripple whose radius increases steadily at 6 cm/s. A toy boat floating on the ponds is 2m east and 1 m north of the spot where the rock is dropped How long does it take for the ripple to reach the boat?

A truck with a wide load, proceeding along a secondary road is approaching a tunnel with a semicircular cross section and a maximum height of 5.25m. If the load is 8m high and 3.5m high will it fit through the tunnel? Explain your answer and include your calculations.

This question I wasnt sure how to start but its worth 7 marks
An error has been made in setting out the square base of a building. the coordinates listed for the corners of the squares are (60,20), (90,120), (-10,140) and (-40,50). Which coordinate pair is incorrect? Explain your choice. What should the coordinates be ? Write a paragraph explaining what the error is, how it was detected, and how it should be corrected

2. I tried these problems out but I am having too much trouble

A rock dropped out into a pond sends out a circular ripple whose radius increases steadily at 6 cm/s. A toy boat floating on the ponds is 2m east and 1 m north of the spot where the rock is dropped How long does it take for the ripple to reach the boat?
Use Pythagoras to find the distance from the point to the boat and divide it by 0.06.

A truck with a wide load, proceeding along a secondary road is approaching a tunnel with a semicircular cross section and a maximum height of 5.25m. If the load is 8m wide and 3.5m high will it fit through the tunnel? Explain your answer and include your calculations.
You have a semicircle of radius 21/4. $y=\sqrt{(\frac{21}{4})^{2}-x^{2}}$. The truck will center the middle of the road. So, what do you get if x=4?. Is it more than 3.5?. If it is, the truck won't make it.

This question I wasnt sure how to start but its worth 7 marks
An error has been made in setting out the square base of a building. the coordinates listed for the corners of the squares are (60,20), (90,120), (-10,140) and (-40,50). Which coordinate pair is incorrect? Explain your choice. What should the coordinates be ? Write a paragraph explaining what the error is, how it was detected, and how it should be corrected

Plot the points and you can easily see if it's a square or not.

3. Thanks for the help but for the first question could I just convert the 2m and 1m in to cm units and then it Would be 300 cm and divide that by 6cm/s to get my answer? I am still confused on how you would find the distance from the point to the boat and for the second question I have no idea how to solve it using that formula where exactly did you get 21/4 and the thing is I am in grade 10 math and I know for sure we dont solve math problems using the formula you did and our teacher never taught us that way is there any other way to do that problem for gr 10 math?And for the last question how would I plot the points since there isnt enough room to plot that on graph paper?

4. #1. Yeah, you can convert anyway you wish. $\sqrt{2^{2}+1^{2}}=\sqrt{5}=2.23 \;\ meters$. Where did you get the 300?.

#2. 21/4=5.25. I just converted to a fraction. Have no idea how to solve?. Nothing to solve. Just plug in x=4. Is the y value greater than 3.5?. Draw a picture of the truck going through the tunnel. It'll help.
How is your teacher expecting you to tackle it?.

#3. Divide everything by 10. 100. Whatever, to make it fit.

BTW, please use a little punctuation. It's difficult to tell where one topic ends and the next begins.

5. Thanks alot I understand it more now
For #1 I know how you got 2.23 meters so after you get 2.23 meters would you convert 2.23 meters into centimetres and get 223 centimetres? And after the conversion would you just divide 223 by 6cm/s and get 37.16s?

For #2 I dont have a graphing calculator only a scientific. I think the teacher wants us to do it by like using the distance formulas and the other formulas for analytical geometry. I am sure she doesnt want us to do it the way you mentioned it we never did problems like that since its grade 10 math.

For #3 I just divide by 10 to make it fit. By dividing by 10 that to make it smaller that wont change the overall outcome of the qeustions right?

6. #1 is good.

#2 it's just Pythagoras, as the other problems are.

#3. No, it won't change anything, just scale.

7. Thanks alot for #1and3 but what exactly do I have to do for #2 that one is still confusing for me

8. Poor TH1, I think you're just confusing yourself. It's not bad. Here's a graph.

It's a Pythagoras, just like the rock and boat from #1.

The equation of a circle with radius 5.25 is $y=\sqrt{5.25^{2}-x^{2}}$. See the truck going thorugh the tunnel?. Since it is 8 feet wide, it is 4 feet on either side of the center(y-axis). By Pythagoras, $\sqrt{5.25^{2}-4^{2}}=3.4$. That's a hair less than 3.5, ain't it?.
At a point 4 feet over, the tunnel is 3.4 feet high. But the truck is 3.5 feet high. Oops.

9. For the units its not feet its meters. For the last line of your sentence I know that the truck will be too high to go through the tunnel but I really didnt understand this part
" At a point 4ft over the tunnel is 3.4 feet high. But the truck is 3.5 feet high"

And I would have to show my calculations so I could show a drawing. But I think I would need more calculations since they would want me to show everything and how I got it.

10. 3.
Let $A(60,20),B(90,120),C(-10,140),D(-40,50)$ be the corners of the quadrilater.
We'll find the slopes of the sides.
$\displaystyle m_{AB}=\frac{y_B-y_A}{x_B-x_A}=\frac{10}{3}$
$\displaystyle m_{BC}=-\frac{1}{5},m_{CD}=3,m_{DA}=-\frac{3}{10}$
We have $m_{AB}\cdot m_{DA}=-1\Rightarrow AB\perp DA$, so the quadrilater has only one right angle.
We'll find the lengths of the sides:
$AB=\sqrt{(x_A-x_B)^2+(y_A-y_B)^2}=10\sqrt{109}$
$BC=10\sqrt{104},CD=10\sqrt{90},DA=10\sqrt{109}$
We have $AB=DA$, so $A,B,D$ can be the corners of a square, but $C$ can't be.

Now, let's make the corection.
Let $C(x,y)$. First of all, the quadrilater must be a parallelogram, so, the point of intersection of diagonals must be the midpoint of each diagonal.
Then $\displaystyle\frac{x_A+x_C}{2}=\frac{x_B+x_D}{2}$ and $\frac{y_A+y_C}{2}=\frac{y_B+y_D}{2}\Rightarrow$
$\Rightarrow\frac{60+x}{2}=\frac{50}{2}\Rightarrow x=-10$
$\frac{20+y}{2}=\frac{170}{2}\Rightarrow y=150$
Thus $C(-10,150)$

11. For the part where you find the length of AB and BC can you show me how you got that since I tried it and couldnt get it