use implicit differentiation to find the slope of the tangent line to the curve of x^2/3+y^2/3=4 at the point (-1,3sqrt3)
taking the first
2/3 x dx + 2/3 ydy=0
dy/dx=-x/y at the point, 1/(3sqrt3)
((x^2)/3)+((y^2)/3)=4
d/dx 2x/3 + d/dx 2y/3 (y')= 0
y'= -(2x/3)-(2y/3)
at the point (-1,3sqrt3)
y'=(-2(-1)/3)-(2(3sqrt3)/3)=(2/3)-((6sqrt3)/3)=(2/3)-2sqrt3
Answer is (2/3)-2sqrt3 correct?
Is is x^(2/3) or is it (x^2)/3 ?
If it was the latter you could multiply both sides by 3 to get rid of any fractions straight away.
Brittany, y' = -(2x/3) / (2y/3) = -x/y
not y' = -(2x/3)-(2y/3)
it is x^(2/3)
The problem is (x^(2/3))+(y^(2/3))=4
I'm confused on the answer
I got the same answer as Brittany, I just want to make sure if it is correct or not
ok let's see...
x^(2/3) + y^(2/3) = 4
differentiating we get
(2/3)*(x^(-1/3)) + (2/3)*(y^(-1/3)).dy/dx = 0
divide across by (2/3) to get
x^(-1/3) + y^(-1/3).dy/dx = 0
Then replace the negative power with 1/ instead. e.g. x^-1 = 1/x
1/x^(1/3) + (1/y^(1/3)).dy/dx = 0
(1/y^(1/3)).dy/dx = -1/x^(1/3)
dy/dx = -y^(1/3)/x^(1/3) = -(y/x)^1/3
pt (1, 3sqrt3)
dy/dx = -(3sqrt3)^1/3
Bit nasty to be honest.
of course, you can simplify that last result to just √3, since it is (3^(3/2))^(1/3) = 3^(1/2)
Also, you dropped/gained a minus sign in there somewhere. The point is in QII, where the slope is positive.
To find the slope of the tangent line to the curve of \(x^{\frac{2}{3}} + y^{\frac{2}{3}} = 4\) at the point \((-1, 3\sqrt{3})\), we can use implicit differentiation. Implicit differentiation allows us to differentiate both sides of an equation with respect to the appropriate variables while treating one variable as an implicit function of the other.
Let's follow the steps to find the derivative with respect to \(x\):
1. Differentiate both sides of the equation with respect to \(x\).
\( \frac{d}{dx} (x^{\frac{2}{3}} + y^{\frac{2}{3}}) = \frac{d}{dx} 4 \)
2. Use the chain rule to differentiate.
\( \frac{2}{3}x^{-\frac{1}{3}} + \frac{2}{3}y^{-\frac{1}{3}} \cdot \frac{dy}{dx} = 0 \)
3. Move \(\frac{2}{3}x^{-\frac{1}{3}}\) to the other side.
\( \frac{2}{3}y^{-\frac{1}{3}} \cdot \frac{dy}{dx} = -\frac{2}{3}x^{-\frac{1}{3}} \)
4. Simplify the equation.
\( \frac{dy}{dx} = \frac{-x^{-\frac{1}{3}}}{y^{-\frac{1}{3}}} \)
Now, substitute the coordinates of the given point \((-1, 3\sqrt{3})\) into \(\frac{dy}{dx}\) to find the slope at that point.
\( \frac{dy}{dx} = \frac{-(-1)^{-\frac{1}{3}}}{(3\sqrt{3})^{-\frac{1}{3}}} \)
Simplifying further, remember that \((-1)^{-\frac{1}{3}}\) is \( -1 \) since we are not including the imaginary root.
\( \frac{dy}{dx} = \frac{-1}{(3\sqrt{3})^{-\frac{1}{3}}} \)
To simplify the denominator, we can rationalize the cube root by multiplying both numerator and denominator by \( 3\sqrt{3} \).
\( \frac{dy}{dx} = \frac{-1}{(3\sqrt{3})^{-\frac{1}{3}}} \cdot \frac{(3\sqrt{3})^{\frac{1}{3}}}{(3\sqrt{3})^{\frac{1}{3}}} \)
\( \frac{dy}{dx} = \frac{-3\sqrt{3}}{3} \)
\( \frac{dy}{dx} = -\sqrt{3} \)
Therefore, the slope of the tangent line to the curve at the point \((-1, 3\sqrt{3})\) is \( -\sqrt{3} \).