# tangent to the curve

• Jan 17th 2008, 02:52 AM
Oranges&Lemons
tangent to the curve
i think i'm making a silly mistake somewhere, because i should be able to do this!

A curve has the equation y= 3x+11 (all square rooted)
The point P on the curve has x-coordinate 3

Show that the tangent to the curve at P has the equation 3x-4root5y + 31=0

any help would be greatly appreciated, i cant seem to see where ive gone wrong!
• Jan 17th 2008, 03:02 AM
mr fantastic
Quote:

Originally Posted by Oranges&Lemons
i think i'm making a silly mistake somewhere, because i should be able to do this!

A curve has the equation y= 3x+11 (all square rooted)
The point P on the curve has x-coordinate 3

Show that the tangent to the curve at P has the equation 3x-4root5y + 31=0

any help would be greatly appreciated, i cant seem to see where ive gone wrong!

Since you know you've gone wrong, you're obviously getting an answer different to the given one. It would be useful to see your working, that is, to see how you got your answer. Did you follow this basic approach:

1. $\frac{dy}{dx} = \frac{3}{2 \sqrt{3x + 11}}$ therefore the gradient of the tangent to the curve at P is given by m = .....

2. A point on the tangent to the curve at P is (3, 20).

3. The equation of any line (including a tangent) has the general form $y - y_1 = m(x - x_1)$.

4. Conclude that the given answer of $3x - 4 \sqrt{5} y +$ 31 = 0 is wrong - that the 31 should be $80 \sqrt{5} - 9$ ....
• Jan 17th 2008, 03:14 AM
Oranges&Lemons
Quote:

Originally Posted by mr fantastic
Since you know you've gone wrong, you're obviously getting an answer different to the given one. It would be useful to see your working, that is, to see how you got your answer. Did you follow this basic approach:

1. $\frac{dy}{dx} = \frac{3}{2 \sqrt{3x + 11}}$ therefore the gradient of the tangent to the curve at P is given by m = .....

2. A point on the tangent to the curve at P is (3, 20).

3. The equation of any line (including a tangent) has the general form $y - y_1 = m(x - x_1)$.

4. Conclude that the given answer of $3x - 4 \sqrt{5} y +$ 31 = 0 is wrong - that the 31 should be $80 \sqrt{5} - 9$ ....

ok...its the differentiating ive done wrong! i differentiated it as...

0.5(3x+11)to the power -0.5 then times 3....

so 1.5(3x+11)to the power -0.5
ive used the wrong rule havent i...
• Jan 17th 2008, 03:16 AM
Oranges&Lemons
hang on are you saying the answer given on the paper is wrong? its not -31 on the end?
• Jan 17th 2008, 03:25 AM
mr fantastic
Quote:

Originally Posted by Oranges&Lemons
ok...its the differentiating ive done wrong! i differentiated it as...

0.5(3x+11)to the power -0.5 then times 3....

so 1.5(3x+11)to the power -0.5
ive used the wrong rule havent i...

Your derivative is correct. 1.5 is the same as 3/2. Raising to -0.5 is the same as taking the reciprocal square root.
• Jan 17th 2008, 03:28 AM
mr fantastic
Quote:

Originally Posted by Oranges&Lemons
hang on are you saying the answer given on the paper is wrong? its not -31 on the end?

Wait a moment. Let me just go back and read exactly what I wrote ......

Quote:

Originally Posted by mr fantastic
[snip]
4. Conclude that the given answer of $3x - 4 \sqrt{5} y +$ 31 = 0 is wrong - that the 31 should be $80 \sqrt{5} - 9$ ....

Quote:

Originally Posted by Oranges&Lemons
hang on are you saying the answer given on the paper is wrong? its not -31 on the end?

Yep, that's what I said alright.
WAIT WAIT WAIT. WHAT I SAID IS WRONG. LET ME GO BACK AND EDIT MY WORK.
• Jan 17th 2008, 03:30 AM
Oranges&Lemons
it's just clicked! thanks.
• Jan 17th 2008, 03:34 AM
mr fantastic
This is what I should have written:

Quote:

Originally Posted by mr fantastic on an alternative world-line
Since you know you've gone wrong, you're obviously getting an answer different to the given one. It would be useful to see your working, that is, to see how you got your answer. Did you follow this basic approach:

1. $\frac{dy}{dx} = \frac{3}{2 \sqrt{3x + 11}}$ therefore the gradient of the tangent to the curve at P is given by m = .....

2. A point on the tangent to the curve at P is (3, $\sqrt{20}$), that is, (3, $2 \sqrt{5}$).

3. The equation of any line (including a tangent) has the general form $y - y_1 = m(x - x_1)$.

4. Conclude that the given answer of $3x - 4 \sqrt{5} y +$ 31 = 0 is correct.

When I (the Mr F on this world-line) was getting the y-coordinate of P in 2., I looked at the function you posted but forgot to read the (all square rooted) bit at the end .... In my defence, an easy mistake of misreading to make ......
• Jan 17th 2008, 03:37 AM
mr fantastic
Quote:

Originally Posted by Oranges&Lemons
it's just clicked! thanks.

So all's well that ends well.
• Jan 17th 2008, 03:39 AM
Oranges&Lemons
it is +31 though because the y coordinate at P is root 20 not just 20 :) thanks.
• Jan 17th 2008, 03:44 AM
mr fantastic
Quote:

Originally Posted by Oranges&Lemons
it is +31 though because the y coordinate at P is root 20 not just 20 :) thanks.

Yep, see post #8. So, where was your mistake?
• Jan 17th 2008, 04:24 AM
Oranges&Lemons
Quote:

Originally Posted by mr fantastic
Yep, see post #8. So, where was your mistake?

oh yes, my bad! didn't see that post sorry. my mistake was some sort of calculator error when i subbed in the 3...it didnt click that 3/2(3x+11) to the power -1/2 is the same as 3/2root3x+11 either.
• Jan 17th 2008, 04:44 AM
Oranges&Lemons
could you help me with this too by any chance?

i need to differentiate this to do a question...

x^(5/2)ln(x/4)
• Jan 17th 2008, 05:04 AM
colby2152
Quote:

Originally Posted by Oranges&Lemons
could you help me with this too by any chance?

i need to differentiate this to do a question...

x^(5/2)ln(x/4)

Use the product rule: $f = zy$, $f' = z'y + y'z$

Let $z=x^{(5/2)}$ and $y=ln(x/4)$.
• Jan 17th 2008, 05:07 AM
Oranges&Lemons
Quote:

Originally Posted by colby2152
Use the product rule: $f = zy$, $f' = z'y + y'z$

Let $z=x^{(5/2)}$ and $y=ln(x/4)$.

of courseeeee. thanks!