Find an equation for the line tangent to the curve at the given point

x^4+y^3=17 at (-3,-4)

Also if you could post the steps it took you to get the answer plz THANKS GUYS!

Results 1 to 3 of 3

- Sep 27th 2007, 05:40 PM #1flewoveruGuest

- Sep 27th 2007, 05:58 PM #2
Differentiate implicity the function, then plug $\displaystyle (x,y)=(-3,-4)$, so $\displaystyle y'$ will be the slope of the tangent line.

Apply the point-slope formula and you're done.

--

I dunno if this is the procedure, I don't remember these things -.-

- Sep 28th 2007, 05:09 AM #3
As Krizalid said:

$\displaystyle x^4+y^3=17$

$\displaystyle 4x^3 + 3y^2 \frac{dy}{dx} = 0$

$\displaystyle \frac{dy}{dx} = - \frac{4x^3}{3y^2}$

So at (-3, -4) the derivative is:

$\displaystyle \frac{dy}{dx} = - \frac{4(-3)^3}{3(-4)^2} = \frac{4 \cdot 27}{3 \cdot 16} = \frac{9}{4}$

Now you know the tangent line at (-3, -4) has a slope of 9/4. Thus

$\displaystyle y = mx + b$

$\displaystyle -4 = \frac{9}{4} (-3) + b$

$\displaystyle b = -4 + \frac{27}{4} = \frac{11}{4}$

So your line is

$\displaystyle y = \frac{9}{4}x + \frac{11}{4}$

-Dan