Exponential function question (1 Viewer)

[Sien]

Find the equation of the normal to the curve y=e^x at the pt where x=3 in exact form.
Step by step explanation would be appreciated I can't tell of the textbook's ans is wrong or if mine is wrong. Thanks :>

 

[enigma_1]

dy/dx = e^x

at the point x=3 sub it into the derivative to get the gradient:

f'(3) = e^3

since we need the normal, find the perpendicular gradient, ie -1/e^3

at the point x=3, sub it into y=e^x to find the y coordinate.
ie y=e^3

:. The point is (3, e^3)

eqn of tangent can be found using the point gradient formula y-y1=m(x-x1)

where x1 = 3 and y1= e^3 and m= -1/e^3

sub it all in:

y- e^3 = -1/e^3 (x-3)

y-e^3 = -x/e^3 + 3/e^3

multiply all terms by e^3

ye^3 - e^6 = -x + 3

:. x + ye^3 - e^6 - 3 = 0

 

[Sien]

dy/dx = e^x

at the point x=3 sub it into the derivative to get the gradient:

f'(3) = e^3

since we need the normal, find the perpendicular gradient, ie -1/e^3

at the point x=3, sub it into y=e^x to find the y coordinate.
ie y=e^3

:. The point is (3, e^3)

eqn of tangent can be found using the point gradient formula y-y1=m(x-x1)

where x1 = 3 and y1= e^3 and m= -1/e^3

sub it all in:

y- e^3 = -1/e^3 (x-3)

y-e^3 = -x/e^3 + 3/e^3

multiply all terms by e^3

ye^3 - e^6 = -x + 3

:. x + ye^3 - e^6 - 3 = 0

Thanks enigma much appreciation XD

 

[Speed6]

Thanks enigma much appreciation XD

Enigma is a legend

 

[enigma_1]

Thanks enigma much appreciation XD

No worries

Enigma is a legend

Speedyyy <3

I'm no legend haha