1. ## Problem Set

I think I figured out the first two questions, but I have three more puzzlers to go. Can anyone help?

3. Verify that the line 4x-y+11+0 is tangent to the curve y=(16/x²)-1. Then determine the point of tangency.

I started this question off by finding the derivative of the curve, and I got y'= -32x(-3), where -3 is the exponent on x. I also isolated for y in the line and I got y=4x+11.

I have no idea where to go from here.

4. For what values of "a" and "b" will the parabola y=x²+ax+b be tangent to the curve y=x²√x+1 at the point (3, 18)?

*Stares blankly at the question.*

5. Find the AREA of the triangle formed by the three points on the curve y=[x²-1 / x²+1]² where the tangent lines are horizontal.

We haven't done anything like this before! We never even learned the equations on how to do this! All we have discussed in this course (which started a couple weeks ago) has been limits, derivatives, power rule, etc.

2. Hello, NAPA55!

Here's the last one . . .

5. Find the area of the triangle formed by the three points
on the curve: . $y\:=\:\left(\frac{x^2-1}{x^2+1}\right)^2$ where the tangent lines are horizontal.
First, find the derivative (Chain Rule, Quotient Rule, etc.)

. . $y' \;=\;2\left(\frac{x^2-1}{x^2+1}\right)\cdot\frac{(x^2+1)\cdot2x - (x^2-1)\cdot2x}{(x^2+1)^2} \;=\;\frac{8x(x^2-1)}{(x^2+1)^3}$

Tangents are horizontal where $y' = 0$
. . So we have: . $8x(x-1)(x+1) \:=\:0\quad\Rightarrow\quad x \;=\;0,\:\pm1$

When $x = 0\!:\;\;y \:=\:\left(\frac{0-1}{0+1}\right)^2 \:=\:1\quad\Rightarrow\quad (0,\,1)$

When $x = 1\!:\;\;y \:=\:\left(\frac{1-1}{1+1}\right)^2 \:=\:0\quad\Rightarrow\quad (1,\,0)$

When $x = \text{-}1\!:\;\;y \:=\:\left(\frac{1-1}{1+1}\right)^2\:=\:0\quad\Rightarrow\quad (\text{-}1,\,0)$

And surely you can find the area of this triangle . . .

Code:
                |
*(0,1)
/ | \
/   |   \
/     |     \
- * - - - + - - - * -
(-1,0)     |     (1,0)

3. Thanks!!

4. Originally Posted by NAPA55
...
3. Verify that the line 4x-y+11+0 is tangent to the curve y=(16/x²)-1. Then determine the point of tangency.

I started this question off by finding the derivative of the curve, and I got y'= -32x(-3), where -3 is the exponent on x. I also isolated for y in the line and I got y=4x+11.

I have no idea where to go from here.

...
The slope of the line (m = 4) equals the gradient of the curve for a certain x-value. Therefore you have to determine a x-value so that

$4 = -\frac{32}{x^3}~\implies~4x^3 = -32~\implies~x^3 = -8 ~\implies~ x = -2$

Plug in this value into the equation of the line (or the curve; it doesn't matter) and you'll get the tangent point. I've got T(-2, 3)

5. Originally Posted by NAPA55
...
4. For what values of "a" and "b" will the parabola y=x²+ax+b be tangent to the curve y=x²√(x+1) at the point (3, 18)?

*Stares blankly at the question.*
...

At the tangent point both curves must have the same gradient:

$f(x)=x^2 \sqrt{x + 1} ~\implies~ f'(x)=\frac{x(5x+4)}{2\sqrt{x+1}}$ And therefore:

$f'(3) = \frac{57}{4}$ must equal $p'(x)=2x + a$ with x = 3, that means $p'(3) = 6+a$

The tangent point must be located on the parabola. With both conditions you get a system of equations:

$\left|\begin{array}{l}18 = 9+3a+b\\ \frac{57}{4} =2 \cdot 3 + a\end{array}\right.$

From the 2nd equation you get: $a = \frac{33}{4}$ . Plug in this value into the 1rst equation to calculate b:

$18=9+3 \cdot \left(\frac{33}{4} \right) +b~\implies~ b = - \frac{63}{4}$

The equation of the parabola becomes:

$p: y = x^2 + \left(\frac{33}{4} \right) x - \frac{63}{4}$

EDIT: I changed all results according to the new wording of the question.

6. The root applies to the whole (x+1).

For what values of "a" and "b" will the parabola y=x²+ax+b be tangent to the curve y=x²√(x+1) at the point (3, 18)?

7. And for #3, how do I show the "verifying" part?

8. Originally Posted by NAPA55
The root applies to the whole (x+1).

For what values of "a" and "b" will the parabola y=x²+ax+b be tangent to the curve y=x²√(x+1) at the point (3, 18)?
I've made all necessary changes in my previous post. Have a look there.

9. Originally Posted by NAPA55
And for #3, how do I show the "verifying" part?
All calculations had been done to get one point which belongs to both graphs and where both graphs have the same gradient

Therefore this point is a tangent point and both graphs are tangent curves to each other.

This process is called verifying.(I believe )

10. Originally Posted by earboth
At the tangent point both curves must have the same gradient:

$f(x)=x^2 \sqrt{x + 1} ~\implies~ f'(x)=\frac{x(5x+4)}{2\sqrt{x+1}}$
How did you do that step?

11. Originally Posted by NAPA55
How did you do that step?
use product rule:

$f(x)=x^2 \sqrt{x + 1} ~\implies~ f'(x) = 2x \sqrt{x+1} + x^2 \cdot \frac12 \cdot (x+1)^{-\frac12} =$ $\frac{2 \cdot 2x \sqrt{x+1} \cdot \sqrt{x+1} + x^2}{2\sqrt{x+1}} = \frac{ 4x (x+1) + x^2}{2\sqrt{x+1}} = \frac{x(5x+4)}{2\sqrt{x+1}}$

12. Thanks... stupid me... I didn't realize that until I immediately posted that message and shut off the computer.

13. Thanks.