What is the equation of the tangent to the cu | 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 $y = 2 \cos \left( \frac{\pi}{4} \right) = 2 \left( \frac{1}{\sqrt{2}} \right) = \sqrt{2}$.So,the point of tangency is $\left( \frac{\pi}{4}, \sqrt{2} \right)$.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 passing through $(x_1, y_1)$ with slope $m$ is $y - y_1 = m(x - x_1)$.Substituting the values,we get $y - \sqrt{2} = -\sqrt{2} \left( x - \frac{\pi}{4} \right)$.

What is the equation of the tangent to the curve $y = 2 \cos x$ at $x = \pi / 4$?

Similar Questions

The $x$-coordinate of the point on the curve $y = a\left( {{e^{\frac{x}{a}}} + {e^{ - \frac{x}{a}}}} \right)$ where the tangent is parallel to the $x$-axis is ........

The point on the curve $y = \sqrt{x - 1}$ where the tangent is perpendicular to the line $2x + y - 5 = 0$ is

If $\frac{k}{\alpha^3}$ is the length of the subnormal at any point $P(\alpha, y)$ on the curve $x^2-a^2=\frac{x^2 y^2}{a^2}$,then $k=$

The length of the subnormal to any point on a curve is always constant. Then,the curve is . . . . . . .

The abscissa of the point on the curve $y=a\left(e^{\frac{x}{a}}+e^{-\frac{x}{a}}\right)$ where the tangent is parallel to the $X$-axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module