Question

If $x=\sec\theta-\cos\theta$, $y=\sec^{10}\theta-\cos^{10}\theta$ then $\left(\frac{dy}{dx}\right)^{2}$ is equal to

• A. $100\left(\frac{y^{2}+4}{x^{2}+4}\right)$
• B. $100\left(\frac{y^{2}+4}{x^{2}-4}\right)$
• C. $100\left(\frac{y^{4}-4}{x^{4}+4}\right)$
• D. $100\left(\frac{y^{4}+2}{x^{4}+4}\right)$
• E. $100\left(\frac{y^{4}+4}{x^{4}+2}\right)$

Hint:

Math Tip: Whenever you see parametric equations of the form $u = f(\theta) - \frac{1}{f(\theta)}$ and $v = f(\theta)^n - \frac{1}{f(\theta)^n}$, differentiating and squaring them will almost always require the identity $(A+B)^2 = (A-B)^2 + 4AB$ to convert the terms back into the original $u$ and $v$ variables.

Answer 1

The Correct Option is A

Concept:
Calculus - Parametric Differentiation and Algebraic Identities.
To find $\frac{dy}{dx}$ for parametric equations, use the chain rule: $\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}$.
Useful Identity: $(a+b)^2 = (a-b)^2 + 4ab$.
Step 1: Differentiate $x$ with respect to $\theta$.
Given $x = \sec\theta - \cos\theta$. $$ \frac{dx}{d\theta} = \sec\theta\tan\theta - (-\sin\theta) $$ $$ \frac{dx}{d\theta} = \sec\theta\tan\theta + \sin\theta $$ Factor out $\tan\theta$ by rewriting $\sin\theta$ as $\tan\theta\cos\theta$: $$ \frac{dx}{d\theta} = \tan\theta(\sec\theta + \cos\theta) $$
Step 2: Differentiate $y$ with respect to $\theta$.
Given $y = \sec^{10}\theta - \cos^{10}\theta$. Apply the power and chain rules: $$ \frac{dy}{d\theta} = 10\sec^9\theta(\sec\theta\tan\theta) - 10\cos^9\theta(-\sin\theta) $$ $$ \frac{dy}{d\theta} = 10\sec^{10}\theta\tan\theta + 10\cos^9\theta\sin\theta $$ Factor out $10\tan\theta$ by rewriting $\cos^9\theta\sin\theta$ as $\cos^{10}\theta\tan\theta$: $$ \frac{dy}{d\theta} = 10\tan\theta(\sec^{10}\theta + \cos^{10}\theta) $$
Step 3: Evaluate $\frac{dy}{dx}$.
Divide the derivative of $y$ by the derivative of $x$: $$ \frac{dy}{dx} = \frac{10\tan\theta(\sec^{10}\theta + \cos^{10}\theta)}{\tan\theta(\sec\theta + \cos\theta)} $$ Cancel out the common $\tan\theta$ terms: $$ \frac{dy}{dx} = \frac{10(\sec^{10}\theta + \cos^{10}\theta)}{\sec\theta + \cos\theta} $$
Step 4: Square the derivative to find $\left(\frac{dy}{dx}\right)^2$.
$$ \left(\frac{dy}{dx}\right)^2 = \frac{100(\sec^{10}\theta + \cos^{10}\theta)^2}{(\sec\theta + \cos\theta)^2} $$
Step 5: Apply the algebraic identity to the numerator and denominator.
Use the identity $(a+b)^2 = (a-b)^2 + 4ab$. Note that $\sec\theta \cdot \cos\theta = 1$.

• For the Numerator: $(\sec^{10}\theta + \cos^{10}\theta)^2 = (\sec^{10}\theta - \cos^{10}\theta)^2 + 4(\sec^{10}\theta\cos^{10}\theta)$
  Since $y = \sec^{10}\theta - \cos^{10}\theta$, this becomes $y^2 + 4(1) = y^2 + 4$.
• For the Denominator: $(\sec\theta + \cos\theta)^2 = (\sec\theta - \cos\theta)^2 + 4(\sec\theta\cos\theta)$
  Since $x = \sec\theta - \cos\theta$, this becomes $x^2 + 4(1) = x^2 + 4$.

Step 6: Construct the final equation.
Substitute the substituted identities back into the squared derivative formula: $$ \left(\frac{dy}{dx}\right)^2 = \frac{100(y^2 + 4)}{x^2 + 4} $$