+0  
 
0
1130
5
avatar+1314 
see image attached
http://snag.gy/Rd9Td.jpg
 Apr 4, 2014

Best Answer 

 #5
avatar+109783 
+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.

 Jun 2, 2014
 #1
avatar+109783 
+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?
 Apr 4, 2014
 #2
avatar+30386 
+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)).
 Apr 4, 2014
 #3
avatar+111396 
+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
 Apr 4, 2014
 #4
avatar+1314 
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.

 Jun 1, 2014
 #5
avatar+109783 
+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

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