Finding the Point to Shut Off Engines for a Traveler in Space

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Homework Help Overview

The problem involves a traveler in space moving along the curve defined by y = x^2. The traveler needs to determine the point on this curve at which to shut off the engines to reach a specified point (4, 15) by following the tangent line.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the derivative of the curve and the slope of the tangent line. There are attempts to set up equations relating the slope to the coordinates of the tangent point. Questions arise regarding the correctness of the quadratic equation derived from these slopes.

Discussion Status

Participants are actively engaging with the problem, exploring different approaches to find the point of tangency. Some have identified errors in their calculations and are reconsidering their methods. There is mention of different potential methods to solve the problem, including graphing.

Contextual Notes

Participants note confusion regarding the quadratic equation and its implications for finding the correct point. There is also mention of a distinction between an "easy" and a "hard" method as suggested by a teacher, though specifics are not provided.

Physics1
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Homework Statement



A traveler in space is moving left and right on y = x^2. He shuts off the engines and continues on the tangent line until he reaches point (4, 15). At what point on the curve should he shut off the engines to reach that point?

The Attempt at a Solution



Ok, the derivative is 2x. I know have to solve for the slope of the tangent line.

Rise over run = (15-a^2) / (4 - a)

(a, a^2) is the point to shut off. I get a^2 - 8a + 15 as the quadratic equation but the value 4 from it isn't correct when I plug it into 2x. What am I doing wrong?
 
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What is the general form of a straight line?
 
Hootenanny said:
What is the general form of a straight line?

y = mx + b

I know. I did that but it isn't the correct answer when looking at the graph.

I got y = 8x - 17. It doesn't touch the graph
 
So we now have to equations for the gradient and our point x=a;

[tex]m = 2a \hspace{1cm};\hspace{1cm}m = \frac{15-a^2}{4-a}[/tex]

Can you now solve for a?
 
Hootenanny said:
So we now have to equations for the gradient and our point x=a;

[tex]m = 2a \hspace{1cm};\hspace{1cm}m = \frac{15-a^2}{4-a}[/tex]

Can you now solve for a?

I see what I did wrong now. I did the quadratic wrong because I accidentally put a wrong number in so I had to do the quadratic equation which led to a wrong number. 3 works for a and the point is (3, 9). Isn't there another way to solve this?
 
Physics1 said:
I see what I did wrong now. I did the quadratic wrong because I accidentally put a wrong number in so I had to do the quadratic equation which led to a wrong number. 3 works for a and the point is (3, 9). Isn't there another way to solve this?
Not that I can think of...
 
Hootenanny said:
Not that I can think of...

My teacher said there's an easy way (I'm assuming it's this way) and a hard way.
 
Physics1 said:
My teacher said there's an easy way (I'm assuming it's this way) and a hard way.
The only other option I can think of is graphing it, and with sketching skill that would be the hard way!
 

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