Tangents & Normals to Logs (1 Viewer)

[Mc_Meaney]

Hey guys, Im having trouble with two questions, any help is much appreciated

(i) FInd the equation of the tanjent to the curve y=2+e^3x at the point x=0

(ii) Find the exact gradient of the normal to the curve y=x-e^-x aat the point

where x=2


Thanks in advance

 

[Riviet]

i) dy/dx=3e^3x
When x=0,
dy/dx = 3e⁰
=3
Also when x=0, y=2+e⁰
=2+1
=3
.'. equation of tangent is y-3=3(x-0)
y-3=3x
3x-y+3=0

ii)dy/dx=1+e^-x
When x=2,
dy/dx=1+e^-2
So gradient of tangent is 1+ 1/e²=(e²+1)/e²
.'. gradient of normal is -e²/(e²+1)

 

[pLuvia]

(i)
d/dx(2+e^3x)=3e^3x
At x=0 y=3
dy/dx=3
y-3=3(x-0)
3x-y+3=0

(ii)
d/dx(x-e^-x)=1+e^-x
At x=2
dy/dx=1+e^-2
Gradient to normal m₁m₂=-1
1+e^-2.m₂=-1
m₂=-1/(1+e^-2)

Edit: Mistake and Riviet beat me to it

 

[Mc_Meaney]

Thanks Riv, once again you've sufficiently confused the hell outta me :rofl:

EDIT: Guys, the book doesn't like that answer, its given 3x-y+3= 0 for the (i) and for (ii) its given -e/e^2+1

Confused :'(

 

[pLuvia]

I'm guessing they're wrong, if they rationalise mine and Riviet's answer to (ii) it gives something close but not to their answer, are you sure you typed the answer to (ii) right? I could probably understand it being -e²/(e²+1). Actually if it was this -e²/(e²+1) then yes it's right we just didn't simplify it any furthur, but it would still be considered the right answer.

Ok silly mistake for (i) that should be right now

 

[Mc_Meaney]

I'm guessing they're wrong, if they rationalise mine and Riviet's answer to (ii) it gives something close but not to their answer, are you sure you typed the answer to (ii) right? I could probably understand it being e²/(e²+1)

Hmm, typed it rong it is meant to be what you typed above, just a negative
(-e^2/e^2+1)

yeh so. Lol

 

[Riviet]

Fixed (i) and (ii), I've checked each a few times and pretty sure they're right.

 

[Mc_Meaney]

Just 1 more quick question, because I am so confused

What is the working for 3^x=8 (finding x)

:burn:

 

[Mc_Meaney]

Scrap that, I think I've got it

log(e)8/log(e)3
=1.892....
=1.9 (answer In book)

Does this look right?

 

[Riviet]

log₃(3^x)=log₃8

Edit: yep that's correct.

 

[Mc_Meaney]

THANKS GUYS!!! I owe you big time. I get it now :rofl:

But I am having trouble with 2 more questions then I'll stop asking I promise

(1) Solve: 12 = 10e^0.001t

(2) Log(9)8

(3) d/dx of Inx is like...1/x yeh?

 

[Riviet]

(1) 12/10 = e^0.001t
ln(6/5)=0.001t
t=ln(6/5) / 0.001

(2) log₉8=(log8)/(log9)
Note: you can use any log base (10 or e) as long as you keep it consistent, ie (log₁₀8)/(log₁₀9) or (log_e8)/(log_e9)

(3) Yep, you got it.

 

[sando]

I just started the topic: exponential and logarithmic functions today, so i'm having a bit of trouble...
Can someone plz help with these 2 questions....

1) Find the second derivatice of (e^2x + 1)^7

2) Find the gradient of the tangent to the curve y = e^5x at the point where x = 0

 

[insert-username]

I just started the topic: exponential and logarithmic functions today, so i'm having a bit of trouble...
Can someone plz help with these 2 questions....

1) Find the second derivatice of (e^2x + 1)^7

You need to use the chain rule and the derivative of an exponential here:

d/dx[(e^2x + 1)⁷]

= 7(e^2x + 1)⁶.2e^2x

= 14e^2x.(e^2x + 1)⁶

Then derive again (this time using the product rule and chain rule).

d/dx[14e^2x.(e^2x + 1)⁶]

= 14e^2x.2e^2x.6(e^2x + 1)⁵ + (e^2x + 1)⁶.28e^2x

= 168e^4x(e^2x + 1)⁵ + (e^2x + 1)⁶.28e^2x

= (e^2x + 1)⁵[168e^4x + (e^2x + 1).28e^2x]

2) Find the gradient of the tangent to the curve y = e^5x at the point where x = 0

dy/dx = 5e^5x

Therefore when x = 0, dy/dx = 5 (since anything to the power 0 is one, e^5x = 1)

Therefore the gradient of the tangent when x = 0 is 5.


I_F

 

[sando]

Thanks man....

one more question till i have finished the h/w lol... it would have helped if i didnt miss half the lesson

1) Find the equation of the tangent to the curve y = e^x^2 -3x at the point where x = 0

 

[SoulSearcher]

OK, y = e^x2 - 3x
when x = 0,
y = e⁰ - 3 * 0
y = 1, therefore co-ordinates are (0,1)
y' = 2xe^x2 - 3
when x = 0,
y' = 2 * 0 * e⁰ - 3
y' = -3, therefore grapient of tangent is -3
therefore using the formula y = mx + b, where m is the gradient and b is the y intercept,
y = -3x + 1
therefore the equation of the tangent is 3x + y - 1 = 0

 

[sando]

thats exactly what i got.... damn the stupid book... must have wrong answer

it fustrated me all maths lesson and for an 1hr when i got home

thanks for that... now i now its not just me

 

[Riviet]

1) Find the equation of the tangent to the curve y = e^x^2 -3x at the point where x = 0

dy/dx=e^x2.2x - 3

At x=0,

m=e⁰.0 - 3

=-3

When x=0, y=1-0=1

.'. equation of tangent is:

y-1=-3(x-0)

y-1=-3x

3x+y-1=0

P.S Try not to rely on the book's answers too much.

 

[sando]

Yeah i hav found several times that maths in focus is a pretty shitty txtbook...

i tried this new question on the next exercise... and got stuck

1) Find any stationary points on the curve y = x^2 e^2x and sketch the curve

i got up to the bit... 2xe^2x(1+x) = 0 .... but i don't know how to solve that
just head me in the right direction and then i will be fine to finish it off..
thanks

 

[Riviet]

2x.e^2x(x+1)=0
2x.e^2x=0 or (x+1)=0
x=0, -1

Substitute into original equation to find y-values of each stat point and test the nature of them using either first or second derivative test.