Given a curve whose locus is given by:
$$ ^{1/3} + ^{1/3} = 15$$. Find the value of slope of the curve at [5, 5].
I have worked out like following. $$y^{1/3} = 15 - x^{1/3}$$ $$ y = (15- x^{1/3})^3 $$ So, value of $dy/dx$ at $x=5$, should give the solution. Is this approach the right one?
On
I think that is could be safer to consider the implicit function $$f(x,y)=x^{\frac 13}+y^{\frac 13}-15=0$$ Then, computing the partial derivatives $$\frac {\partial f(x,y)} {\partial x}=\frac 13 x^{-\frac 23}\qquad \text{and} \qquad \frac {\partial f(x,y)} {\partial y}=\frac 13 y^{-\frac 23}$$ Yhen, using the implicit function theorem $$\frac{dy}{dx}=-\frac{\frac {\partial f(x,y)} {\partial x} } {\frac {\partial f(x,y)} {\partial y} }=-\left(\frac y x \right)^{\frac 23}$$
On
Lets take $$x=15^3\cos^6(t), y=15^3\sin^6(t)$$ $$\frac{dy}{dx}=\frac{\sin^5(t)\cos(t)}{-\sin(t)\cos^4(t)}=-\tan^4(t)$$ Slope of line joining $(5,5)$ and point with parameter $t$ on the curve is $$\frac{15^3\sin^6(t)-5}{15^3\cos^6(t)-5}=-\tan^4(t)$$
Solving for $t$ would give tangents from $(5,5)$ on the curve.
Yes, your approach is totally fine. It's going to work here because $y$ can de expressed as an explicit function of $x$. But generally, when it's not possible to get $y$ isolated on one side all by itself and move all the $x$'s over to the other side, you would use something called implicit differentiation:
$$ x^{1/3}+y^{1/3}=15\\ \frac{d}{dx}\left(x^{1/3}\right)+\frac{d}{dx}\left(y^{1/3}\right)=\frac{d}{dx}\left(15\right)\\ \frac{1}{3x^{2/3}}+\frac{1}{3y^{2/3}}\frac{dy}{dx}=0\\ \frac{dy}{dx}=-\frac{3y^{2/3}}{3x^{2/3}}\\ \frac{dy}{dx}=-\left(\frac{y}{x}\right)^{\frac{2}{3}}. $$
Now, just plug in the $x$ and $y$ values of the point that you're interested in. This should give you the value of the slope of the tangent line to the curve at that point. In your case, however, as Fred mentioned in the comments section, the point $(5,5)$ that you're given does not seem to be on the curve because it does not satisfy the original equation. So, you might want to check your problem statement again.