Kuroyuki's Spesh thread

[kinslayer]

You don't need to use cos^2 x + sin^2 x = 1 for this one.

dy/dx = dy/dt * dt/dx = cos t * 1/(1 + sin t). When t = pi/6, dy/dx = sqrt(3)/3 = 1/sqrt(3)

Now you know the gradient, find F(pi/6) = (x(pi/6), y(pi/6)) and you can write down the equation of the tangent line.

[Kuroyuki]

You don't need to use cos^2 x + sin^2 x = 1 for this one.

dy/dx = dy/dt * dt/dx = cos t * 1/(1 + sin t). When t = pi/6, dy/dx = sqrt(3)/3 = 1/sqrt(3)

Now you know the gradient, find F(pi/6) = (x(pi/6), y(pi/6)) and you can write down the equation of the tangent line.

yeah but if I use  cos^2 x + sin^2 x = 1 and implicitly differentiate I don't get dydx as sqrt(3)/3. I'm wondering why :/

[kinslayer]

yeah but if I use  cos^2 x + sin^2 x = 1 and implicitly differentiate I don't get dydx as sqrt(3)/3. I'm wondering why :/

I guess I don't see why you are trying to use cos^2 x + sin^2 x = 1. That trick is helpful to eliminate the t's if you need an equation in y and x only (no parameter), but it doesn't help here, since x^2 does not equal cos^2 t. If you tried to go down that road you'd end up with x^2 + y^2 = (some function of t) and it's a mess.

If the equations were nice, like x(t) = cos t and y(t) = sin t, you could say x^2 + y^2 = 1 and you would be able to use implicit differentiation to get a value for the gradient of the tangent in terms of x and y. But if you are given the point on the curve in terms of t (like you are in this problem), it would still be quicker to use the chain rule: dy/dx = cos t / (-sin t) = -cot t.

You don't need to eliminate the t's in this equation so you can forget about cos^2 x + sin^2 y = 1.

[Kuroyuki]

I guess I don't see why you are trying to use cos^2 x + sin^2 x = 1. That trick is helpful to eliminate the t's if you need an equation in y and x only (no parameter), but it doesn't help here, since x^2 does not equal cos^2 t. If you tried to go down that road you'd end up with x^2 + y^2 = (some function of t) and it's a mess.

If the equations were nice, like x(t) = cos t and y(t) = sin t, you could say x^2 + y^2 = 1 and you would be able to use implicit differentiation to get a value for the gradient of the tangent in terms of x and y. But if you are given the point on the curve in terms of t (like you are in this problem), it would still be quicker to use the chain rule: dy/dx = cos t / (-sin t) = -cot t.

You don't need to eliminate the t's in this equation so you can forget about cos^2 x + sin^2 y = 1.

Ah ok. Thank you!

[Kuroyuki]

Could someone please help me with the last part of this question.
Thanks in advance!

[kinslayer]

The first thing to see is that the maximum angle is made where the shooting trajectory is tangent to the curve.

You found the equations of all tangents to the curve in part b), now you need to find which one of them passes through the player's position at (4,0).

Start with y = -u/(4v)x + 1/v. When y = 0, x = 4, so u = 1. Remember (u,v) is a point on C. Use the equation for C which is x^2/4 + y^2 = 1. Plug in x = 1 and you find that v = y = (+/-) sqrt(3)/2.

Now you know u and v, plug them into the tangent equation, and you will get the two tangents that pass through G. If you take v = sqrt(3)/2 (top of C) then the angle it makes with the x-axis is t where tan t = sqrt(3)/6. Use symmetry to conclude that the maximum (aiming through from the top of C to the bottom of C) is 2t.

[Kuroyuki]

The first thing to see is that the maximum angle is made where the shooting trajectory is tangent to the curve.

You found the equations of all tangents to the curve in part b), now you need to find which one of them passes through the player's position at (4,0).

Start with y = -u/(4v)x + 1/v. When y = 0, x = 4, so u = 1. Remember (u,v) is a point on C. Use the equation for C which is x^2/4 + y^2 = 1. Plug in x = 1 and you find that v = y = (+/-) sqrt(3)/2.

Now you know u and v, plug them into the tangent equation, and you will get the two tangents that pass through G. If you take v = sqrt(3)/2 (top of C) then the angle it makes with the x-axis is t where tan t = sqrt(3)/6. Use symmetry to conclude that the maximum (aiming through from the top of C to the bottom of C) is 2t.

Thanks so much! I understand.