Q.Find the tangent and normal to the following curves at the given points on the curve.

Step 1 (i). $y=x^2-x^4$ at $(1,0)$.

$y'=2x-4x^3\Rightarrow y'(1)=2-4=-2$. Tangent: $y-0=-2(x-1)\Rightarrow 2x+y-2=0$. Normal slope $=\tfrac12$: $y-0=\tfrac12(x-1)\Rightarrow x-2y-1=0$.

Step 2 (ii). $y=x^4+2e^x$ at $(0,2)$.

$y'=4x^3+2e^x\Rightarrow y'(0)=0+2=2$. Tangent: $y-2=2(x-0)\Rightarrow2x-y+2=0$. Normal slope $=-\tfrac12$: $y-2=-\tfrac12x\Rightarrow x+2y-4=0$.

Step 3 (iii). $y=x\sin x$ at $\left(\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$.

$y'=\sin x+x\cos x\Rightarrow y'\!\left(\tfrac{\pi}{2}\right)=1+\tfrac{\pi}{2}(0)=1$. Tangent: $y-\tfrac{\pi}{2}=1\left(x-\tfrac{\pi}{2}\right)\Rightarrow y=x\Rightarrow x-y=0$. Normal slope $=-1$: $y-\tfrac{\pi}{2}=-\left(x-\tfrac{\pi}{2}\right)\Rightarrow x+y-\pi=0$.

Step 4 (iv). $x=\cos t,\ y=2\sin^2t$ at $t=\pi/3$.

$\dfrac{dx}{dt}=-\sin t$, $\dfrac{dy}{dt}=4\sin t\cos t\Rightarrow\dfrac{dy}{dx}=\dfrac{4\sin t\cos t}{-\sin t}=-4\cos t$. At $t=\pi/3$: $\cos\tfrac{\pi}{3}=\tfrac12\Rightarrow$ slope $=-2$. …

Differentiate (implicit/parametric where needed), then apply the tange …

• Forgetting the product rule on $x\sin x$ or the chain rule on $e^x$-type terms …