# Math Help - Differential calculus help/check....

1. ## 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

2. 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:

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

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

Good luck!

3. Hi, thanks for replying okay

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.

4. 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.

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

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

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

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

5. 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?