Find the slope of the tangent to the following curves at the respective given points.

(i) $y=x^4+2x^2-x$ at $x=1$

(ii) $x=a\cos^3t,\ y=b\sin^3t$ at $t=\dfrac{\pi}{2}$.

Question

Find the slope of the tangent to the following curves at the respective given points.

(i) $y=x^4+2x^2-x$ at $x=1$

(ii) $x=a\cos^3t,\ y=b\sin^3t$ at $t=\dfrac{\pi}{2}$.

Accepted Answer

> [!TLDR]
>
> Part (i) is direct differentiation; part (ii) needs the parametric-slope formula $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$, and recognising that an unbounded slope means the tangent is vertical, not "no tangent."

**Step 1 (i). Differentiate $y=x^4+2x^2-x$.**
$y'=4x^3+4x-1$. At $x=1$: $y'(1)=4(1)+4(1)-1=7$.

**Step 2 (ii). Differentiate the parametric curve.**
$x=a\cos^3t\Rightarrow \dfrac{dx}{dt}=-3a\cos^2t\sin t$; $\ y=b\sin^3t\Rightarrow\dfrac{dy}{dt}=3b\sin^2t\cos t$.

**Step 3. Form $\dfrac{dy}{dx}$.**
$$\frac{dy}{dx}=\frac{3b\sin^2t\cos t}{-3a\cos^2t\sin t}=-\frac{b}{a}\cdot\frac{\sin t}{\cos t}=-\frac{b}{a}	an t.$$

**Step 4. Evaluate as $t	o\pi/2$.**
As $t	o\pi/2^-$, $\cos t	o0^+$ so $	an t	o+\infty$, hence $\dfrac{dy}{dx}	o-\infty$ — an infinite (undefined) slope. At $t=\pi/2$, both $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ are individually $0$ (an astroid cusp-type point on the axis), and the curve there is at $\big(a\cos^3	frac{\pi}{2},\,b\sin^3	frac{\pi}{2}\big)=(0,b)$, where the tangent is the **vertical line** $x=0$.
> [!ANSWER]
> (i) The slope at $x=1$ is $7$.
> (ii) At $t=\pi/2$ the slope is infinite — the tangent is the vertical line $x=0$.

Q.Find the slope of the tangent to the following curves at the respective given points.

Concept understanding — Tangents and Normals (Lengths)

Tangents and Normals to a Curve

Q.Find the slope of the tangent to the following curves at the respective given points.

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