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The normal to the curve x=a(cosθ+θ+sinθ), y=a(sinθ-θ cosθ) at any point 'θ' is such that :
- a)it makes a constant angle with the x-axis
- b)it passes through the origin
- c)it is at a constant distance from the origin
- d)none of these
Correct answer is option 'C'. Can you explain this answer?
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Most Upvoted Answer
The normal to the curve x=a(cosθ+θ+sinθ), y=a(sin&th...
T+tsin(t)), y=a(sin(t)-tcos(t)) at the point where t=0 is:
To find the normal, we need to find the derivative of the curve and evaluate it at t=0. We can use the chain rule to find the derivative:
dx/dt = -a(sin(t)+tcos(t))
dy/dt = a(cos(t)-sin(t)-tcos(t))
So the slope of the tangent at t=0 is dy/dx evaluated at t=0:
dy/dx = (dy/dt)/(dx/dt) = (cos(t)-sin(t)-tcos(t))/(-sin(t)-tcos(t))
dy/dx at t=0 is (1-0-0)/(-0-0) = 0
Therefore, the normal to the curve at t=0 is a vertical line, so its equation is x=a.
To find the normal, we need to find the derivative of the curve and evaluate it at t=0. We can use the chain rule to find the derivative:
dx/dt = -a(sin(t)+tcos(t))
dy/dt = a(cos(t)-sin(t)-tcos(t))
So the slope of the tangent at t=0 is dy/dx evaluated at t=0:
dy/dx = (dy/dt)/(dx/dt) = (cos(t)-sin(t)-tcos(t))/(-sin(t)-tcos(t))
dy/dx at t=0 is (1-0-0)/(-0-0) = 0
Therefore, the normal to the curve at t=0 is a vertical line, so its equation is x=a.
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Community Answer
The normal to the curve x=a(cosθ+θ+sinθ), y=a(sin&th...
T+tsint), y=a(sin t-tcost) at the point where t=0 is:
First, we need to find the equation of the curve by eliminating the parameter t. We have:
x=a(cos t+t sin t) = a cos t + a t sin t
y=a(sin t-t cos t) = a sin t - a t cos t
Squaring both equations and adding them, we get:
x^2 + y^2 = a^2 (1+t^2)
Solving for t^2, we get:
t^2 = (x^2 + y^2)/a^2 - 1
Since t=0 at the point where we want to find the normal, we have:
t = sqrt((x^2 + y^2)/a^2 - 1)
Now, we need to find the derivative of the curve with respect to t:
dx/dt = -a sin t + a t cos t
dy/dt = a cos t + a t sin t
At t=0, we have:
dx/dt = 0
dy/dt = a
So the slope of the tangent at t=0 is dy/dx = dy/dt / dx/dt = infinity.
Therefore, the normal is a horizontal line passing through the point (a,0). Its equation is:
y = 0
First, we need to find the equation of the curve by eliminating the parameter t. We have:
x=a(cos t+t sin t) = a cos t + a t sin t
y=a(sin t-t cos t) = a sin t - a t cos t
Squaring both equations and adding them, we get:
x^2 + y^2 = a^2 (1+t^2)
Solving for t^2, we get:
t^2 = (x^2 + y^2)/a^2 - 1
Since t=0 at the point where we want to find the normal, we have:
t = sqrt((x^2 + y^2)/a^2 - 1)
Now, we need to find the derivative of the curve with respect to t:
dx/dt = -a sin t + a t cos t
dy/dt = a cos t + a t sin t
At t=0, we have:
dx/dt = 0
dy/dt = a
So the slope of the tangent at t=0 is dy/dx = dy/dt / dx/dt = infinity.
Therefore, the normal is a horizontal line passing through the point (a,0). Its equation is:
y = 0
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The normal to the curve x=a(cosθ+θ+sinθ), y=a(sinθ-θ cosθ) at any point θ is such that :a)it makes a constant angle with the x-axisb)it passes through the originc)it is at a constant distance from the origind)none of theseCorrect answer is option 'C'. Can you explain this answer?
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