#5**+10 **

Yes Stu, if two lines are perpendicular then the product of their gradients will be -1

$$m_1m_2=-1\\\\

therefore\\\\

m_2=\dfrac{-1}{m_1}$$

The normal to a curve is the name given to a line that is perpendicular to the tangent of the curve at that point.

That is; The normal is perpendicular to the tangent.

Melody Jun 2, 2014

#1**+5 **

Hi Stu,

(a)

the graph y=e^{x} has a very special property. The y value is equal to the gradient of the tangent.

so

dy/dx also equals e^{x}

Reiterating:

If y=e^{x} then y'=e ^{x}

This is your question

f(x)=1-e^{x}

You are asked to find the gradient of the tangent to the curve at the point where it crosses the x axis.

In this case y has been replaced with f(x). [Call it f(x) but treat it the same as if it was y]

Any curve crosses the x axis when y=0 or in this case, f(x)=0 This is the x intercept.

Find the x intercept

0=1-e^{x}

e^{x}=1

x=0

now

f(x)=1-e^{x}

f'(x)=0-e^{x} = -e ^{x}

when x=0

f'(0)=-e^{0} = -1

this is the end of part a.**I haven't done b or c yet.**

Part b is missing the point - do you know what it is supposed to be?

(a)

the graph y=e

so

dy/dx also equals e

Reiterating:

If y=e

This is your question

f(x)=1-e

You are asked to find the gradient of the tangent to the curve at the point where it crosses the x axis.

In this case y has been replaced with f(x). [Call it f(x) but treat it the same as if it was y]

Any curve crosses the x axis when y=0 or in this case, f(x)=0 This is the x intercept.

Find the x intercept

0=1-e

e

x=0

now

f(x)=1-e

f'(x)=0-e

when x=0

f'(0)=-e

this is the end of part a.

Part b is missing the point - do you know what it is supposed to be?

Melody Apr 4, 2014

#2**+5 **

...

Part b is missing the point - do you know what it is supposed to be?[/quote]

I think it means the point at which f(x) crosses the x axis (i.e. the point (0, 0)).

Part b is missing the point - do you know what it is supposed to be?[/quote]

I think it means the point at which f(x) crosses the x axis (i.e. the point (0, 0)).

Alan Apr 4, 2014

#3**+7 **

Part (b)

We want to find a tangent line to f(x) = 1 - e^x where it crosses the x axis. As posted earlier, this point is (0,0)

So, the derivative of this unction gives the slope (gradient) at any point on the curve. So, the derivative is given by f'(x) = -e^x. So, the slope of the tangent line at this point is just -(e^0) = -1. So, using the point-slope form we have:

y - 0 = (-1)(x - 0)

y = -x

So, in Part (c), we want to find the normal line to this. Then the slope of the normal line is just 1. And using the point-slope form in a similar manner we have:

y - 0 = (1)(x - 0)

y = x

We want to find a tangent line to f(x) = 1 - e^x where it crosses the x axis. As posted earlier, this point is (0,0)

So, the derivative of this unction gives the slope (gradient) at any point on the curve. So, the derivative is given by f'(x) = -e^x. So, the slope of the tangent line at this point is just -(e^0) = -1. So, using the point-slope form we have:

y - 0 = (-1)(x - 0)

y = -x

So, in Part (c), we want to find the normal line to this. Then the slope of the normal line is just 1. And using the point-slope form in a similar manner we have:

y - 0 = (1)(x - 0)

y = x

CPhill Apr 4, 2014

#4**0 **

Is the perpendicular tangent to a line -1/m? if so why is it also the normal tangent? This has me confused, or im ilinformed.

Stu Jun 1, 2014

#5**+10 **

Best Answer

Yes Stu, if two lines are perpendicular then the product of their gradients will be -1

$$m_1m_2=-1\\\\

therefore\\\\

m_2=\dfrac{-1}{m_1}$$

The normal to a curve is the name given to a line that is perpendicular to the tangent of the curve at that point.

That is; The normal is perpendicular to the tangent.

Melody Jun 2, 2014