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

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SUMMARY

The discussion centers on determining the optimal point for a traveler moving along the curve defined by y = x² to shut off his engines and reach the point (4, 15). The traveler must find the point (a, a²) where the slope of the tangent line, given by the derivative 2a, matches the slope calculated from the coordinates of the target point. The correct solution is found at a = 3, yielding the point (3, 9). The conversation also touches on alternative methods for solving the problem, including graphing.

PREREQUISITES
  • Understanding of calculus concepts, specifically derivatives and tangent lines
  • Familiarity with quadratic equations and their solutions
  • Knowledge of the general form of a straight line equation (y = mx + b)
  • Basic graphing skills for visualizing curves and tangents
NEXT STEPS
  • Study the application of derivatives in finding tangent lines to curves
  • Learn how to solve quadratic equations using the quadratic formula
  • Explore graphical methods for solving equations and visualizing functions
  • Investigate alternative approaches to optimization problems in calculus
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Students studying calculus, educators teaching optimization problems, and anyone interested in the application of derivatives in real-world scenarios.

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;

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

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

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

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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