CSSA trial 2012 question 14b (1 Viewer)

[Isabellatanl]

Does anyone know what's the solution?




A man M walks along a pier, represented by the positive y-axis, pulling on a boat B(x,y) by a rope of length L. The man is initially at the origin O and the boat is initially on the x-axis, L metres from O, The man keeps the rope taut and the path followed by the boat is such that the rope is always tangent to the curve tracing its path.




i) let the path followed by the boat be the graph of the function y=f(x). By considering the gradient of the line MB, show that dy/dx = -surd(L^2-x^2) /x.

 

[Myans]

X^2 + Y^2 = L^2
Y = rt(L^2 - X^2)

Gradient of MB = Rise/Run = Y/-X = rt(L^2 - X^2)/-X = -rt(L^2-X^2)/X [Negative X because from the original point (x,y), the X value is decreasing]

 

[Isabellatanl]

How do u know gradient of MB is y/x? Since M(0,y1) and B(x,y)? The gradient depends on coordinate M also? I mean it depends on y1 also? And then for the Pythagoras theorem, can u explain tat?

 

[Myans]

How do u know gradient of MB is y/x?

Gradient of any line is Rise/Run --> The height/the width. Since the question specifies to consider this, your attention should immediately focus on that.

for the Pythagoras theorem, can u explain tat?

In a triangle, a^2 + b^2 = c^2 where a and b are adjacent sides, and c is the hypotenuse. In this diagram. you have the the hypotenuse L (length of string) with sides y and x. Therefore x^2 + y^2 = L^2
y^2 = L^2 - x^2 [Minus x^2 from both sides]
y = rt (L^2 - x^2) [Square root both sides] [Take only positive value as we are working with distances, which are always > 0]

 

[Isabellatanl]

Thanks for ur help. I understood alr.