Vedora · Class 12 · Question 1 grading

Question 1

1/5 marks

Problem

If $x = a\cos^{3}t$ , $y = a\sin^{3}t$ then $\dfrac{dy}{dx}$ at $t = \dfrac{3\pi}{4}$ is

(A) $-1$

(B) $-\dfrac{1}{\sqrt{3}}$

(D) $1$

Student answer

Student work

$x=a\cos^{3}t\qquad y=a\sin^{3}t$

$\dfrac{dx}{dt}=3a\cos^{2}t\sin t\qquad \dfrac{dy}{dt}=3a\sin^{2}t\cos t$

$\frac{dy}{dx}=\frac{3a\sin^{2}t\cos t}{3a\cos^{2}t\sin t}=\tan t$

at $t=\dfrac{3\pi}{4}$ , $\dfrac{dy}{dx}=\tan\dfrac{3\pi}{4}=-1$

option (a) is correct

Strengths

Used parametric differentiation $\left(\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}\right)$ .
Computed $\dfrac{dy}{dt}=3a\sin^{2}t\cos t$ correctly.
Evaluated $\tan\left(\dfrac{3\pi}{4}\right)$ correctly (though it came from the wrong intermediate).

Marking rationale

The student used the correct parametric method. They computed $\dfrac{dy}{dt}$ correctly but missed the negative sign in $\dfrac{dx}{dt}$ . Because the error occurs early, the formation of $\dfrac{dy}{dx}$ and the final evaluation depend on that wrong value and earn no credit. Partial credit was given for the correct $\dfrac{dy}{dt}$ .

What to improve

Differentiation of trigonometric functions Moderate

Missed the negative sign when differentiating $\cos^{3}t$ (forgot $\dfrac{d}{dt}[\cos t]=-\sin t$ ).

From your work: $\dfrac{dx}{dt}=3a\cos^{2}t\sin t$

Try this: Review the chain rule and derivatives of sin and cos. Practice differentiating $(\cos t)^{n}$ and $(\sin t)^{n}$ and check signs carefully.