Explanation For finding the equation y = f ( x ) , we have to solve both parametric equations and have to eliminate the variable t . (a) Derivative of a function in parametric form can be found by differentiating both functions with respect to the parametric variable and then by dividing them d y d x = d y d θ d x d θ = y ' ( θ ) x ' ( θ ) (b) After finding the equation , we have to differentiate the equation with respect to x, and have to find f ' ( x ) . Given: x = cos θ , y = cos θ + sin 2 θ Formula Used: (1) Differentiation of x n is d ( x n ) d x = n x n − 1 (2) d y d x = d y d θ d x d θ = y ' ( θ ) x ' ( θ ) (3) Differentiation d sin θ d θ = cos θ , d cos θ d θ = − sin θ Calculation: For finding f ( x ) , use x = cos θ , y = cos θ + sin 2 θ . x = cos θ cos θ = x sin 2 θ = 1 − x 2 Putting the value of sin θ , cos θ , in y . y = cos θ + sin 2 θ y = x + 1 − x 2 y = − x 2 + x + 1 ð f ( x ) = − x 2 + x + 1 Part (a) x = cos θ , y = cos θ + sin 2 θ x = cos θ d x d θ = − sin θ y = cos θ + sin 2

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Chapter 12.1, Problem 62E

To determine

Find an equation $y = f (x)$ for the parametric curve and compute $\frac{d y}{d x}$ in two ways for

$x = \cos θ, y = \cos θ + \sin^{2} θ$

(a) Using $\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{y' (θ)}{x' (θ)}$ .

(b) By differentiation of $f (x)$ .

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Chapter 12.1, Problem 61E

Chapter 12.1, Problem 63E

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Chapter 12 Solutions

CALCULUS 4E (LL) W/ ACHIEVE ACCESS

Ch. 12.1 - Prob. 1PQ Ch. 12.1 - Prob. 2PQ Ch. 12.1 - Prob. 3PQ Ch. 12.1 - Prob. 4PQ Ch. 12.1 - Prob. 5PQ Ch. 12.1 - Prob. 6PQ Ch. 12.1 - Prob. 7PQ Ch. 12.1 - Prob. 1E Ch. 12.1 - Prob. 2E Ch. 12.1 - Prob. 3E

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Find an equation y = f ( x ) for the parametric curve and compute d y d x in two ways for x = cos θ , y = cos θ + sin 2 θ (a) Using d y d x = d y d θ d x d θ = y ' ( θ ) x ' ( θ ) . (b) By differentiation of f ( x ) .

Solution Summary: The author explains how to find the equation y = f (x) by solving both parametric equations and eliminating the variable t.

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Find an equation $y = f (x)$ for the parametric curve and compute $\frac{d y}{d x}$ in two ways for

$x = \cos θ, y = \cos θ + \sin^{2} θ$

(a) Using $\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{y' (θ)}{x' (θ)}$ .

(b) By differentiation of $f (x)$ .

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Chapter 12 Solutions

Find an equation y = f ( x ) for the parametric curve and compute d y d x in two ways for x = cos θ , y = cos θ + sin 2 θ (a) Using d y d x = d y d θ d x d θ = y ' ( θ ) x ' ( θ ) . (b) By differentiation of f ( x ) .

Solution Summary: The author explains how to find the equation y = f (x) by solving both parametric equations and eliminating the variable t.

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Find an equation y=f(x) for the parametric curve and compute dydx in two ways for x=cosθ,y=cosθ+sin2θ (a) Using dydx=dydθdxdθ=y'(θ)x'(θ). (b) By differentiation of f(x).

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Chapter 12 Solutions

Find an equation $y = f (x)$ for the parametric curve and compute $\frac{d y}{d x}$ in two ways for

$x = \cos θ, y = \cos θ + \sin^{2} θ$

(a) Using $\frac{d y}{d x} = \frac{\frac{d y}{d θ}}{\frac{d x}{d θ}} = \frac{y' (θ)}{x' (θ)}$ .

(b) By differentiation of $f (x)$ .