# Finding the equation of tangent and normal

• March 5th 2009, 07:15 AM
db5vry
Finding the equation of tangent and normal
I have the curve C with equation $3x^{3/2} - \frac{32}{x}$

First need to find equation of the tangent to C at the point where x = 4, and then the equation of the normal to C at the point where x = 4.

All of the questions on my past exam papers are relatively simple however this one seems to be more challenging.

To tell you my working thus far I differentiated the curve to get $4.5x^{1/2} + 32x^{-2}$ and rearranged it into the form of
$4.5x^{1/2} + \frac{32}{x^2}$

And noted that if x was equal to 4, then I would have
$4.5(4)^{1/2} + \frac{32}{4^2}$ which can be written as 9 + 2, which is 11.

I'm not sure what that information is telling me, let alone if it is correct or not. Could you please help?
• March 5th 2009, 07:22 AM
e^(i*pi)
Quote:

Originally Posted by db5vry
I have the curve C with equation $3x^{3/2} - \frac{32}{x}$

First need to find equation of the tangent to C at the point where x = 4, and then the equation of the normal to C at the point where x = 4.

All of the questions on my past exam papers are relatively simple however this one seems to be more challenging.

To tell you my working thus far I differentiated the curve to get $4.5x^{1/2} + 32x^{-2}$ and rearranged it into the form of
$4.5x^{1/2} + \frac{32}{x^2}$

And noted that if x was equal to 4, then I would have
$4.5(4)^{1/2} + \frac{32}{4^2}$ which can be written as 9 + 2, which is 11.

I'm not sure what that information is telling me, let alone if it is correct or not. Could you please help?

You're correct so far, now you have the gradient of the tangent. To find the co-ordinate substitute x=4 into your original equation
$3(4)^{3/2} - \frac{32}{4} = 24 - 8 = 16$

which gives the point (4,16) and gradient 11. From this you can use the equation of a line:
$y - y_1 = m(x-x_1)$ where $(x_1,y_1) = (4,16)$ and m is the gradient of 11.

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As we know the normal passes through the same point we can use the coordinates (4,16) as above. Remember that because the normal is perpendicular to the tangent the products of their gradients must be -1:

$m_1m_2 = -1$ = $11m_2 = -1$ and so $m_2 = -1/11$

You can then use the same equation as above putting m = -1/11 instead of 11.