[kinglrb]

I would really appreciate help with this question that am currently doing for revision.

A curve has equation y= (3x+11)^1/2
The point P on the curve has x-coordinate 3.
(a) Show that the tangent to the curve at P has the equation
3x - 4root5 y + 31 = 0

So what I think I should do is differentiate so I get:
3/2(3x+11)^-1/2

then am stuck

Any help appreciated thanks!

[Jackus]

The differentiation is correct, multiplying by 1/2 and the differentiate which is 3 so 3/2.

Then plug in your x value at p to your dy/dx to give you a value for the gradient of the tangent at p.

Put your x value at p into the original equation to get a y value for that point too.

Then use your x value, y value and gradient to find out c in y=mx+c.

Now you've got everything - just rearrange it into the form they want it.

Sorry if that's not clear, makes sense in my head!

[laura_beth]

find gradient then use:
y-(y value) = gradient(x-xvalue)

[kinglrb]

Originally Posted by Jackus

The differentiation is correct, multiplying by 1/2 and the differentiate which is 3 so 3/2.

Then plug in your x value at p to your dy/dx to give you a value for the gradient of the tangent at p.

Put your x value at p into the original equation to get a y value for that point too.

Then use your x value, y value and gradient to find out c in y=mx+c.

Now you've got everything - just rearrange it into the form they want it.

Sorry if that's not clear, makes sense in my head!

Ahh brillaint my friend, it's very clear just worked it out and it got me the right answer. Thank you!

[OTRamble]

Originally Posted by laura_beth

find gradient then use:
y-(y value) = gradient(x-xvalue)

Can I ask you to explain this, as I instinctively used the y=mx+c such as what Jackus said, however in the mark schemes they used your equation. It confuses me and I don't get it...etc...

[CubeDude]

Originally Posted by OTRamble

Can I ask you to explain this, as I instinctively used the y=mx+c such as what Jackus said, however in the mark schemes they used your equation. It confuses me and I don't get it...etc...

This equation is a useful definition of the gradient, i.e.

Where the delta represents 'change' in the value: .

so



Rearranging gives:


Then simply insert a known point (X,Y), to find the equation of a two dimensional line.

Hope that helps

[OTRamble]

Originally Posted by CubeDude

This equation is a useful definition of the gradient, i.e.

Where the delta represents 'change' in the value: .

so



Rearranging gives:


Then simply insert a known point (X,Y), to find the equation of a two dimensional line.

Hope that helps

Ahhh... yes it does. Thanks!