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CUET PG MCA Previous Year Questions (PYQs)
CUET PG MCA Tangent And Normal PYQ
CUET PG MCA PYQ
Go to Discussion
CUET PG MCA Previous Year PYQCUET PG MCA CUET 2024 PYQ
$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{e^t(\sin t + \cos t)}{e^t(\cos t - \sin t)} = \frac{\sin t + \cos t}{\cos t - \sin t}$$ At \( t = \frac{\pi}{4} \),
$$\sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$$ So, $$\frac{dy}{dx} = \frac{\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}} = \frac{\sqrt{2}}{0}$$ The slope is undefined, which means the tangent is vertical.
Final Answer: The angle with the x-axis is $$\boxed{90^\circ}$$
If the parametric equation of a curve is given by $x=e^t cost$ and $y=e^t sint$ then the tangent to the curve at the point $t=\frac{\pi}{4}$ makes the angle with the axis of x is
Go to Discussion
CUET PG MCA Previous Year PYQCUET PG MCA CUET 2024 PYQ
Solution
Given parametric equations:
$$x = e^t \cos t,\quad y = e^t \sin t$$ To find the angle of the tangent at \( t = \frac{\pi}{4} \), compute the slope:$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{e^t(\sin t + \cos t)}{e^t(\cos t - \sin t)} = \frac{\sin t + \cos t}{\cos t - \sin t}$$ At \( t = \frac{\pi}{4} \),
$$\sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$$ So, $$\frac{dy}{dx} = \frac{\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}} = \frac{\sqrt{2}}{0}$$ The slope is undefined, which means the tangent is vertical.
Final Answer: The angle with the x-axis is $$\boxed{90^\circ}$$
CUET PG MCA
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CUET PG MCA
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and More.
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