C3 help on exp and log

Hi stuck on this question. Please can you explain what to do in each step as im weak on this chapter.

The points P and Q lie on the curve with equation y = e^1/2x. The x coordinates of P and Q are ln4 and ln16 respectively.

a) Find an equation for the line PQ.
b) Show that this line passes through the origin O.
c) Calculate the length, to 3s.f, of the line segment PQ

Thanks for your help - i will give pos rep

Regards, Asad

Reply 1

danniella

4

Hello!

All you have to do is substitute the x-co-ordinates into the equation y = e^{0.5 x} to get the x-co-ordinates of P and Q.

(for P:
y = e^{0.5 ln 4}
y = e^{ln 4^0.5}
y = e^{ln 2}
y = 2

So P= (ln 4, 2)

Do the same thing for Q. You get Q=(ln 16, 4) )

From then on you have two points, P and Q,
and you can find the gradient of the line between them
Gradient = (change in y/ change in x),
then use y-y₁ = m(x - x₁ ) to find the equation of the line.

When you get the equation of the line, put x=0 in the equation and you'll end up finding that y=0 as well. This shows that the line goes through (0,0) which is the origin.

Then to find the length of PQ, use the formula
length= root {(x₁ - x₂)² + (y₁ - y₂)²}

love danniella

Reply 2

generalebriety

16

danniella

Hello!

All you have to do is substitute the x-co-ordinates into the equation y = e^{0.5 x} to get the x-co-ordinates of P and Q.

(for P:
y = e^{0.5 ln 4}
y = e^{ln 4^0.5}
y = e^{ln 2}
y = 2

So P= (ln 4, 2)

Do the same thing for Q. You get Q=(ln 16, 4) )

From then on you have two points, P and Q,
and you can find the gradient of the line between them
Gradient = (change in y/ change in x),
then use y-y₁ = m(x - x₁ ) to find the equation of the line.

When you get the equation of the line, put x=0 in the equation and you'll end up finding that y=0 as well. This shows that the line goes through (0,0) which is the origin.

Then to find the length of PQ, use the formula
length= root {(x₁ - x₂)² + (y₁ - y₂)²}

love danniella

You know, for a GCSE student, you scare me. :biggrin:

Edit: I get 2.43 for the line length.

Reply 3

danniella

4

generalebriety

You know, for a GCSE student, you scare me. :biggrin:

Oh, come *on*. I'm barely five feet tall and weigh around seven stone, how could I *possibly* be scary? :p:

(Edited to add that I got y = 2x/(ln 4) for the equation of the line)

Your mind Danniella.

If you're rather bored and feel like an interesting extension, you can consider the following:

Unparseable latex formula:

Arc Length = \bigint^b_a{\sqrt{1 + (\frac{dy}{dx})^2}}\ dx

So in your case, Arc Length =

Unparseable latex formula:

\bigint^{\ln{16}}_{\ln{4}}{\sqrt{1 + \frac{1}{4}e^x} \ dx

Which is not a particularly lovely integral, and one that I can't do quickly... maybe I'll come back and edit my post.

Bah! I can't do it.

Reply 6

Kolya

17

DoMakeSayThink

So in your case, Arc Length =

Unparseable latex formula:

\bigint^{\ln{16}}_{\ln{4}}{\sqrt{1 + \frac{1}{4}e^x} \ dx

Hmm

Unparseable latex formula:

\bigint^{\ln{16}}_{\ln{4}}{\sqrt{1 + \frac{1}{4}e^x} \ dx

Let

w=1+\frac{1}{4}e^x \\ e^x = 4w-4 \\ x= ln(4w-4) \\ \frac{dx}{dw} = \frac{4}{4w-4} = \frac{1}{w-1}

Now:

Unparseable latex formula:

\bigint{w^{\frac{1}{2}} \frac{dx}{dw}} dw \\ = \bigint{w^{\frac{1}{2}} \frac{1}{w-1}} dw \\ = \bigint{\frac{w^{\frac{1}{2}}}{w-1}} dw

Let

z=w^{\frac{1}{2}} \\ w = z^2 \\ \frac{dw}{dz} = 2z

Now:

Unparseable latex formula:

\bigint{\frac{w^{\frac{1}{2}}}{w-1}} dw \\ = \bigint{\frac{z}{z^2-1}} \frac{dw}{dz} dz \\ = \bigint{\frac{w^{\frac{1}{2}}}{w-1}} dw \\ = \bigint{\frac{z}{z^2-1}}(2z) dz \\ = \frac{2z^2}{z^2-1} dz

Turning into a proper fraction:

Unparseable latex formula:

\bigint{\frac{2z^2}{z^2-1}} dz = \bigint{2+\frac{2}{z^2-1}} dz

Now partial fractions:

Unparseable latex formula:

\bigint{2+\frac{2}{z^2-1}} dz = \bigint{2+\frac{2}{(z-1)(z+1)}} dz = \bigint{2+ \frac{1}{z-1} - \frac{1}{z+1} dz

Giving us