Help with 45° Angle Equation Solving

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To solve for the 45° angle in the context of the equations provided, the relationship between x and y must be established using the equation tan(45°) = y/x, which simplifies to x = y. Given the equations x = 10 + 2t and y = 6 + 4t, substituting yields t = 2. After finding t, the value of (secθ)^2(dθ/dt) can be calculated, resulting in 1/14 per second. It is important to ensure that the calculator is set to the correct mode, as trigonometric derivatives apply only when angles are in radians.
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i can't think of any equation to plug the 45 degrees angle pls help... i tried

Dttanθ=y/x

(secθ)^2)(dθ/dt)=[x(dy/dt)-y(dx/dt)]/x^2

but i don't know the value of x when tanθ=45 degrees... do i need to set my calcu to radians or just in degrees mode then i get tanθ=y/x ---> tan45=y/x

1=y/x
x=y
 
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but i don't know the value of x when tanθ=45 degrees
.
First, it's θ=45 not tanθ=45. Second, if θ=45 what is the relationship between x and y? Using this relationship and x = 10 + 2t, y = 6 + 4t, can you find x and y when θ=45?

i can't think of any equation to plug the 45 degrees angle pls help
How about putting it in for that secθ you have in
(secθ)^2)(dθ/dt)=[x(dy/dt)-y(dx/dt)]/x^2?
 
awww sorry hehe mistyped. it's tan45 hehe:rolleyes:
i came up with this solution
x=10+2t
y=6+4t

tan45=(6+4t)/(10+2t) (deg mode)
1=(6+4t)/(10+2t)
10+2t=6+4t
t=2

substituting in (secθ)^2)(dθ/dt)=[x(dy/dt)-y(dx/dt)]/x^2
i got ((cos45)^2)/7 per sec or 1/14 per sec
is this correct? or should i set my calculator in radian mode?
 
d(sin x)/dx = cos x and d(cos x)/dx= -sin x ONLY if x is in radians. Those formulas are not valid if the angle is not in radians.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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