# math

how do i do this.

find the equation for the lines that are tangent and normal to the curve y= (square root of 2)(cosx)at the point (pi/4, 1)

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3. 👁 72
1. y = √2 cosx
I usually check if the given point is actually on the line
LS = 1
RS = √2 cos(pi/4) = √2(1/√2) = 1 = LS

dy/dx = -√2(sinx)

at (pi/4,1)
dy/dx = -√2(1/√2) = -1

So the slope of the tangent is -1
and the slope of the normal is +1

Take it from there,
every student taking Calculus surely has to know how to find the equation of a line given the slope and a point.

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posted by Reiny
2. use product rule, then substitute value of x in dy/dx. the obtain then find equation of tangent y=mx+c.

For normal,

find gradient of normal from equation and once again obtain normal equation in y=mx+c form

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posted by Salman

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