cambridge maths exercise 7.3 q3 (1 Viewer)

[JamesGoh]

Ok, the solutions to this exercise made the assumption that

tan B = (dy/dt) / (dx/dt)

in order to solve the equation.

Im not seeing the logic behind why the author did this ?

wouldn't it just as be correct to say that tan B = y/x at point t = T/4?

 

[FrankXie]

gradient of tangent; so tan B=(dy)/(dx); by chain rule or implicit differentiation: (dy)/(dx)=(dy/dt)/(dx/dt)

 

[braintic]

Ok, the solutions to this exercise made the assumption that

tan B = (dy/dt) / (dx/dt)

in order to solve the equation.

Im not seeing the logic behind why the author did this ?

wouldn't it just as be correct to say that tan B = y/x at point t = T/4?

I haven't looked at the question. But y/x is the gradient of the line joining the point to the origin, NOT the gradient of the tangent (which is what gives the direction of travel).