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C3 calculus

Hi everyone,

I've just done a question but my answer is wrong. Checked it several times and still can't get the answer. Put it through the calculator and still can't get the values to equal what is in the book. Could anyone tell me where I've gone wrong?

Q: For y=10x−e3x y=10x - e^{3x} find the coordinates of the stationary point and determine it's nature.

My working:
dydx=10−3e3x⇒d2ydx2=−9e3x \frac{dy}{dx}=10-3e^{3x} \Rightarrow \frac{d^{2}y}{dx^{2}}= -9e^{3x}

To find the x coordinate

dydx=0⇒10=3e3x⇒e3x=103⇒13ln103=x \frac{dy}{dx} = 0 \Rightarrow 10=3e^{3x} \Rightarrow e^{3x}= \frac{10}{3} \Rightarrow \frac{1}{3}ln\frac{10}{3}=x. My x coordinate is right, as it is in the answer book.

My y coordinate is wrong though. This was my working, using the original equation and subbing in x:
y=10(13ln103)−e3x y=10(\frac{1}{3}ln\frac{10}{3}) - e^{3x} As I've already found e3x=103e^{3x} = \frac{10}{3}, y is y=(103ln103)−103 y=(\frac{10}{3}ln\frac{10}{3}) - \frac{10}{3}

So my coordinates are: (13ln103,103ln103−103)(\frac{1}{3}ln\frac{10}{3}, \frac{10}{3}ln\frac{10}{3} - \frac{10}{3})

The book says: (13ln103,103ln[103−1])(\frac{1}{3}ln\frac{10}{3}, \frac{10}{3}ln[\frac{10}{3} - 1])

Can anyone tell me what I've done wrong?
Thanks
:smile:
Reply 1
Avatar for Munrot07
Munrot07
18
Original post by marcsaccount
Hi everyone,

I've just done a question but my answer is wrong. Checked it several times and still can't get the answer. Put it through the calculator and still can't get the values to equal what is in the book. Could anyone tell me where I've gone wrong?

Q: For y=10x−e3x y=10x - e^{3x} find the coordinates of the stationary point and determine it's nature.

My working:
dydx=10−3e3x⇒d2ydx2=−9e3x \frac{dy}{dx}=10-3e^{3x} \Rightarrow \frac{d^{2}y}{dx^{2}}= -9e^{3x}

To find the x coordinate

dydx=0⇒10=3e3x⇒e3x=103⇒13ln103=x \frac{dy}{dx} = 0 \Rightarrow 10=3e^{3x} \Rightarrow e^{3x}= \frac{10}{3} \Rightarrow \frac{1}{3}ln\frac{10}{3}=x. My x coordinate is right, as it is in the answer book.

My y coordinate is wrong though. This was my working, using the original equation and subbing in x:
y=10(13ln103)−e3x y=10(\frac{1}{3}ln\frac{10}{3}) - e^{3x} As I've already found e3x=103e^{3x} = \frac{10}{3}, y is y=(103ln103)−103 y=(\frac{10}{3}ln\frac{10}{3}) - \frac{10}{3}

So my coordinates are: (13ln103,103ln103−103)(\frac{1}{3}ln\frac{10}{3}, \frac{10}{3}ln\frac{10}{3} - \frac{10}{3})

The book says: (13ln103,103ln[103−1])(\frac{1}{3}ln\frac{10}{3}, \frac{10}{3}ln[\frac{10}{3} - 1])

Can anyone tell me what I've done wrong?
Thanks
:smile:


The answer is 10/3(ln(10/3) - 1) which when expanded is 10/3ln(10/3) - 10/3

So I think the book may have made a mistake with the brackets
Reply 2
Avatar for marcsaccount
marcsaccount
OP
5
Original post by Munrot07
The answer is 10/3(ln(10/3) - 1) which when expanded is 10/3ln(10/3) - 10/3

So I think the book may have made a mistake with the brackets


Thanks Munrot07 :biggrin:
Reply 3
Avatar for physicsftw
physicsftw
ye, i agree the book's wrong. i got the same answer as you and the stationary point is a maximum.

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