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Differentiation with ln

Hi, I have a seriously cr*p maths teacher and I'm kinda struggling with these questions plus its from the solomon press! :frown:

:-

Equation y= 5e^x - 3lnx has a tangent to the curve at the point Pwith x co ordinate 1
a) Show that the tangent at P has equation y = (5e - 3)x + 3

The curve with equation y= 2x - 2 - lnx passes through the point A(1,0). The tangent to the curve at A crosses the y-axis at B and the normal to the curve at A crosses the y axis at C.
a) Find an equation for the tangent to the curve at A.
b) Show that the midpoint of BC is the origin - huh? :confused:

The curve has a minimum point at D
c) Show that the y coordinate of D is ln2-1

The equation y=e^x passes through the point P(p,e^p). Given that the tangent to the curve at P passes through the origin and that the normal to the curve at P meets the x axis at Q,
a) show that p=1
b) show that the area of triangle OPQ, where O is the origin, is ½e(1+e²).

The curve with the equation y= 4 - e^x meets the y axis at thw point P and the x axis at the point Q.
a) Find an equation for the normal to the curve at P.
b) Find an equation for the tangent the the curve at W.
The normal to the curve at P meets the tangent to the curve at Q at the point R. The x coordinate of R is aln2+b, where a and b are rational constants.
c)Show that a= 8/5
d)Find the value of b.

Reply 1

TomX

I'll answer the first one:

Let x = 1.
y = 5e - 3ln1 = 5e

dy/dx = 5e^x - 3/x (Basic differentiation (i.e. no product or quotient rule, reply if you're stuck on this)

Since x = 1:
dy/dx = 5e - 3

Therefore, since the equation of a straight line is:
y = mx + c
y = (5e - 3)x + c

Now, you know (1, 5e) is a point on the tangent already so substitute x = 1, y = 5e into the equation of the tangent:
5e = 5e - 3 + c
0 = -3 + c

Therefore:
c = 3

Therefore:
y = (5e - 3)x + 3

Reply 2

Da Mouse

can anyone else help me with the last 3 please? :frown:

Reply 3

AlphaNumeric

y = 2x-2-\ln x

\frac{dy}{dx} = 2 - \frac{1}{x}

A = (1,0) Therefore tangent gradient is

m = 2 - (1/1) = 1

Therefore equation is

y-y_{0}=m(x-x_{0})

which is

y-0 = 1(x-1)

y = x-1

b)x=0 => y = -1, so B = (0,-1). A bit of thought tells you C = (0,1).
Midpoint of B and C is (0,0).

c)

\frac{dy}{dx} = 2 - \frac{1}{x}

\frac{dy}{dx} = 0

implies

x = \frac{1}{2}

which implies

y = 1-2-ln(1/2) = -1+\ln 2 = \ln 2 - 1

Reply 4

Da Mouse

I'm still struggling when it comes to e^x pleeeeeeeaaaaaaase someone help me

Reply 5

twisterman

In what way are you struggling? Im crappy at stuff like this - I do know that e^x differentiates to e^x (stays the same) if thats any help? :smile:

Reply 6

AlphaNumeric

y = e^{x}

\frac{dy}{dx} = e^{x}

(p,e^{p})

the equation of the tangent would be

y-e^{p}=e^{p}(x-p)

This passes through (0,0), so when x=0, y=0

0-e^{p} = e^{p}(0-p) \quad \Rightarrow \quad -pe^{p}=-e^{p} \quad \Rightarrow \quad p=1

So P is at (1,e)

If gradient of tangent at P is e, then gradient of normal is

-e^{-1}

. Equation of normal is

y-e=-\frac{1}{e}(x-1)

This passes through the x axis at Q, where y=0.

-e = -\frac{1}{e}(x-1)

x = e^{2}+1

.

So the points on the triangle are (0,0), (1,e) and

(1+e^{2},0)

The base of the triangle is

1+e^{2}

in length and the triangle is e high. Total area is

\frac{1}{2}(1+e^{2})e

For the last question you should be able to use the method I used in this question. Find the gradient of the curve using

\frac{dy}{dx}

, then the equations of the tangent and normals and then where they cross the axes.

Reply 7

Da Mouse

AlphaNumeric

y = e^{x}

\frac{dy}{dx} = e^{x}

(p,e^{p})

the equation of the tangent would be

y-e^{p}=e^{p}(x-p)

This passes through (0,0), so when x=0, y=0

0-e^{p} = e^{p}(0-p) \quad \Rightarrow \quad -pe^{p}=-e^{p} \quad \Rightarrow \quad p=1

So P is at (1,e)

If gradient of tangent at P is e, then gradient of normal is

-e^{-1}

. Equation of normal is

y-e=-\frac{1}{e}(x-1)

This passes through the x axis at Q, where y=0.

-e = -\frac{1}{e}(x-1)

x = e^{2}+1

.

So the points on the triangle are (0,0), (1,e) and

(1+e^{2},0)

The base of the triangle is

1+e^{2}

in length and the triangle is e high. Total area is

\frac{1}{2}(1+e^{2})e

For the last question you should be able to use the method I used in this question. Find the gradient of the curve using

\frac{dy}{dx}

, then the equations of the tangent and normals and then where they cross the axes.

erm the part saying

If gradient of tangent at P is e, then gradient of normal is-----. Equation of normal is This passes through the x axis at Q, where y=0. so --------

isnt showing up on my computer - could you retype the calculation on that part using the normal keys please??

Reply 8

Malik

gradient of normal= -e^-1
equation of normal is y-e=-1/e(x-1)
At Q y=0
-e=-1/e(x-1)
-e²=-x+1
so x=1+e²

So the points on the triangle are (0,0), (1,e) and (1+e²,0)

The base of the triangle is 1+e² in length and the triangle is e high. Total area is (1/2)(1+e²)e

For the last question you should be able to use the method I used in this question. Find the gradient of the curve using dy/dx, then the equations of the tangent and normals and then where they cross the axes.

I dont see some of it clearly either but I think thats what Alpha intended to say.

PS Surbiton High :eek:

just up the road :rolleyes:

Reply 9

Da Mouse

WHAT? I removed that I go to surbiton high 3 days ago! how did you find that out?!?!?!?!

Reply 10

Da Mouse

Thanx guys - I think I just needed reminding about a few things. Its not as complecated as I thought it was. But there is ONE more question I'm kinda stuck on. It asks:-
Curve y=f(x) where f:x ---> 9x^4 - 16lnx, x>0
Given that the set of values of x for which f(x) is a decreasing function of x is 0<x(less or equal to)k, find the exact value of k
The only thing i could think of to do so far is - dy/dx = 36x^3 - 16/x :s:

Reply 11

AlphaNumeric

y is a decreasing function if dy/dx < 0.

You've worked out what dy/dx is. It is less than 0 for a small range, before becoming ( and staying) positive. The value x=k is the point where it goes from being negative to positive, in other words dy/dx = 0 at x=k. Can you see how to solve this?

Reply 12

Dekota

Da Mouse

dy/dx = 36x^3 - 16/x :s:

sub dy/dx = 0...

0 = 36x^3 - 16/x
0 = 1/x(36x^4 - 16)
0 = 36x^4 - 16
16 = 36x^4
±(16/36)^¼ = x
x = ±((16/36)^½)^½ = ±√(4/6) = ±2/√6

Reply 13

MichaelJWilliams

I think you're on the right track - just try 'dy/dx < 0 when f(x) is decreasing' and you should be able to find critical values of x.

Edit: You wait 20 minutes for a response, and then three come along at once. Typical