# Differential calculus help/check....

• Nov 24th 2009, 11:02 AM
playpwnsu
Differential calculus help/check....
Hi I have some questions, I'm not sure if I did them correctly..

1.
a.) The Cissoid of Diocles is a classical curve given by the equation
y^2 = x^3/2-x
Find the slope of the tangent line at the point (1,1)

b.) Consider the point (3,-4) on the circle centred at the origin. Find the equation of the tangent line at this point

Okay this is what I did
y^2 = x^3/2-x
= 2y y' = x^3/2-x
= y' = (1/2y)(x^3)/(2-x)
Sub the point (1,1) into the equation I get
m = 1/2
Then use the point slope equation I get y = 1/2x +1/2???

b.) x^2 + y^2 = 25
2x + 2y y' = 0
y' = -2x-2y
sub the points (3,-4) in I get m = 2
then use point slope equation I get y = 2x-10....

Someone taught me how to do the implicit differentiation for y', thats what I did above not sure If thats how you do it though...........help would be appreciated it :)

Also one more question

2. Find the constants c and K such that the functions f(x) = e^cx + k satisfies the differential equation
3f''(x) + 13f'(x) = 10f(x)

What I did was...
f(x) = e^cx + k
f'(x) = ce^cx
f''(x) = c^2e^cx
Sub these into the equation.
3(c^2e^cx) + 13(ce^cx) = 10(e^cx + k)
3c^2e^cx + 13ce^cx - 10e^cx = 10k
e^cx[3c^2 + 13c - 10] = 10k/e^cx
....then im not sure what to do anymore....
any help would be appreciated :), thanks guys (Talking)
• Nov 24th 2009, 11:49 AM
apcalculus
your implicit differentiation (first step) is incorrect as you haven't differentiated the right hand side.

Use the quotient rule for the right hand side.

The result should be:

$\displaystyle y'=\frac{3 x^2 - x^3}{(-2 + x)^2 y}$

If you show your work here I'll be happy to check.

Good luck!
• Nov 24th 2009, 12:17 PM
playpwnsu

I'm guessing you are replying to Part a of the first question

because I'm not sure if i did the implicit differentiation correctly, and based on your result...

y' = 3x^2 - x^3/(-2 + x)^2y)

I did
y' = (1/2y)(x^3/2-x)
= f'g-g'f/g2
= (3x^2)(2-x) - (-1)(x^3)/(2-x)^2
= 6x^2 - 2x^3/(2-x)^2(2y)
Then simplifly 2
= 3x^2 - x^3/(2-x)^2y

my answer is slightly different from yours though...in the denominator it is 2-x instead of -2+x.
• Nov 24th 2009, 12:30 PM
apcalculus
You're right... it's the same answer, if you pull out a negative sign, then square it from the bottom, you can see that the results are the same.

$\displaystyle y^2 = \frac{x^3}{2-x}$

$\displaystyle 2yy' = \frac{(2-x) 3x^2 - (-) x^3}{(2-x)^2}$

$\displaystyle 2yy' =\frac{6x^2-2x^3}{(2-x)^2}$

$\displaystyle y' =\frac{3x^2-x^3}{y(2-x)^2}$
• Nov 24th 2009, 12:35 PM
playpwnsu
Okay, thanks, so in this case I can plug in the points and find the equation? what about part b?

is there anything I need to change, anything I did wrong?