Find the equation of tangent and normal to the curve y = 2e-x÷3 where the curve meets the Y-axis
Answers
Step-by-step explanation:
Given y=2
3
−x
,
dx
dy
=ln(2)(2
3
−x
)
dx
d
(
3
−x
) Applying exponential functional rule,
=−
3.2
x/3
ln2
The point where at the curve meets y axis, we substitute x=0 in the first equation gives y=1
The slope `m' of the curve at the point (0,1) is given as −
3
ln(2)
.
The equation of tangent at (x
1
,y
1
) is
(y−y
1
)=m(x−x
1
)
The equation of normal at (x
1
,y
1
) is
(y−y
1
)=−
m
1
(x−x
1
), −1/m is the slope of normal
the equation at the points (0,1) for slope −
3
ln(2)
.
y−1=−
3
ln(2)
x (Equation of tangent)
y−1=
ln(2)
3
x (Equation of normal)