Equation of the normal to the curve.

[mac117]

Hello everyone. 

 

I'm revising for my exams this may and I have some troubles with part (b) of this question:

 

"A curve is such that dy/dx = 4x + 1/(x+1)² for x>0 . The curve passes through the point (1/2, 5/6). "

 

a) Find the equation of the curve.

 

My result is :

 

y = 2x²+ 1/(x+1) +1 

 

b) "Find the equation of the normal to the curve at the point where x=1"

 

Do I put the x value into this equation to find the y-co ordinate?(When calculated, y= 5/2) I don't know what to do. Do I take the derivative and calculate it that way? 

 

Any help/suggestion would be appreciated!

[Vioh]

Hello everyone. 

I'm revising for my exams this may and I have some troubles with part (b) of this question:

"A curve is such that dy/dx = 4x + 1/(x+1)² for x>0 . The curve passes through the point (1/2, 5/6). "

a) Find the equation of the curve.

My result is :

y = 2x²+ 1/(x+1) +1 

b) "Find the equation of the normal to the curve at the point where x=1"

Do I put the x value into this equation to find the y-co ordinate?(When calculated, y= 5/2) I don't know what to do. Do I take the derivative and calculate it that way? 

Any help/suggestion would be appreciated!

 

I think you've made a little typo. Your result in (a) should be y = 2x²â€“ 1/(x+1) + 1

 

Now for (b), the first thing that you must realize is that the normal to the curve at a particular point always makes a 90 degrees angle with the tangent line to the curve at that point. This means that the slope of the normal is the negative reciprocal of the slope of the tangent line. And I'm sure you know that the slope of the tangent line at x=1 is basically equal to the derivative of the function at that point. So substituting x=1 into dy/dx, the slope of the tangent line should be 17/4. Hence the slope of the normal is –4/17 (which is the negative reciprocal).

 

The equation describing the normal line is thus: y = mx + c (where m = –4/17). You also know that the normal line must pass through the point (1, 5/2). So all we need to do now is to solve for 'c'

y = mx + c = 5/2 = –4/17 * 1 + c ==> c = 5/2 + 4/17 = 93/34

 

So the equation for the normal is: y = (–4/17)x + 93/34.

[mac117]

 

Hello everyone. 

I'm revising for my exams this may and I have some troubles with part (b) of this question:

"A curve is such that dy/dx = 4x + 1/(x+1)² for x>0 . The curve passes through the point (1/2, 5/6). "

a) Find the equation of the curve.

My result is :

y = 2x²+ 1/(x+1) +1 

b) "Find the equation of the normal to the curve at the point where x=1"

Do I put the x value into this equation to find the y-co ordinate?(When calculated, y= 5/2) I don't know what to do. Do I take the derivative and calculate it that way? 

Any help/suggestion would be appreciated!

 

I think you've made a little typo. Your result in (a) should be y = 2x²â€“ 1/(x+1) + 1

 

Now for (b), the first thing that you must realize is that the normal to the curve at a particular point always makes a 90 degrees angle with the tangent line to the curve at that point. This means that the slope of the normal is the negative reciprocal of the slope of the tangent line. And I'm sure you know that the slope of the tangent line at x=1 is basically equal to the derivative of the function at that point. So substituting x=1 into dy/dx, the slope of the tangent line should be 17/4. Hence the slope of the normal is –4/17 (which is the negative reciprocal).

 

The equation describing the normal line is thus: y = mx + c (where m = –4/17). You also know that the normal line must pass through the point (1, 5/2). So all we need to do now is to solve for 'c'

y = mx + c = 5/2 = –4/17 * 1 + c ==> c = 5/2 + 4/17 = 93/34

 

So the equation for the normal is: y = (–4/17)x + 93/34.

 

 

Thank you very much! 

 

Yes, it was a typo. I did mean -1/(x+1)^^.

 

Finally understood the whole concept. Appreciate your help