The equation of the tangent to the curve y = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the curve $y = 2 \cos x$.At $x = \frac{\pi}{4}$,the value of $y$ is $y = 2 \cos \left( \frac{\pi}{4} \right) = 2 \left( \frac{1}{\sqrt{2}} \

Vedclass · Accepted Answer

$y - \sqrt{2} = -\sqrt{2} \left( x - \frac{\pi}{4} \right)$

Vedclass · Answer

(C) Given the curve $y = 2 \cos x$.At $x = \frac{\pi}{4}$,the value of $y$ is $y = 2 \cos \left( \frac{\pi}{4} \right) = 2 \left( \frac{1}{\sqrt{2}} \right) = \sqrt{2}$.Now,find the derivative $\frac{dy}{dx} = \frac{d}{dx} (2 \cos x) = -2 \sin x$.The slope of the tangent at $x = \frac{\pi}{4}$ is $m = \left( \frac{dy}{dx} \right)_{x = \pi/4} = -2 \sin \left( \frac{\pi}{4} \right) = -2 \left( \frac{1}{\sqrt{2}} \right) = -\sqrt{2}$.The equation of the tangent line at point $(x_1, y_1) = \left( \frac{\pi}{4}, \sqrt{2} \right)$ with slope $m = -\sqrt{2}$ is given by $y - y_1 = m(x - x_1)$.Substituting the values,we get $y - \sqrt{2} = -\sqrt{2} \left( x - \frac{\pi}{4} \right)$.

The equation of the tangent to the curve $y = 2 \cos x$ at $x = \frac{\pi}{4}$ is

Similar Questions

The number of points on the curve $y=54x^5-135x^4-70x^3+180x^2+210x$ at which the normal lines are parallel to $x+90y+2=0$ is:

If the tangent drawn at the point $(x_1, y_1)$,where $x_1, y_1 \in \mathbb{N}$,on the curve $y = x^4 - 2x^3 + x^2 + 5x$ passes through the origin,then $x_1 + y_1 =$

If a variable tangent to the curve $x^2y = c^3$ makes intercepts $a$ and $b$ on the $x$-axis and $y$-axis respectively,then the value of $a^2b$ is

Show that the normal at any point $\theta$ to the curve $x=a \cos \theta+a \theta \sin \theta, y=a \sin \theta-a \theta \cos \theta$ is at a constant distance from the origin.

If the two curves $x=y^2$ and $xy=k$ cut each other at right angles, then a possible value of $8k^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module