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 t d x d t = y ' ( t ) x ' ( t ) (b) After finding the equation , we have to differentiate the equation with respect to x, and have to find f ' ( x ) . Given:- c ( t ) = ( 1 2 t , 1 4 t 2 − t ) i.e. x = t 2 , y = 1 4 t 2 − t 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 t d x d t = y ' ( t ) x ' ( t ) Calculation:- For finding f ( x ) , use x = t 2 , y = 1 4 t 2 − t . x = t 2 t = 2 x Putting the value of t , in y . y = t 2 4 − t y = ( 2 x ) 2 4 − 2 x y = 4 x 2 4 − 2 x y = x 2 − 2 x ð f ( x ) = x 2 − 2 x Part (a) x = t 2 , y = 1 4 t 2 − t

4th Edition

ISBN: 9781319050733

Author: Rogawski

Publisher: MAC HIGHER

See similar textbooks

Concept explainers

Minimization

In mathematics, traditional optimization problems are typically expressed in terms of minimization. When we talk about minimizing or maximizing a function, we refer to the maximum and minimum possible values of that function. This can be expressed in ter…

Maxima and Minima

The extreme points of a function are the maximum and the minimum points of the function. A maximum is attained when the function takes the maximum value and a minimum is attained when the function takes the minimum value.

Derivatives

A derivative means a change. Geometrically it can be represented as a line with some steepness. Imagine climbing a mountain which is very steep and 500 meters high. Is it easier to climb? Definitely not! Suppose walking on the road for 500 meters. Which …

Concavity

In calculus, concavity is a descriptor of mathematics that tells about the shape of the graph. It is the parameter that helps to estimate the maximum and minimum value of any of the functions and the concave nature using the graphical method. We use the …

Videos

Question

Chapter 12.1, Problem 60E

To determine

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

$c (t) = (\frac{1}{2} t, \frac{1}{4} t^{2} - t)$

(a) Using $\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{y' (t)}{x' (t)}$ .

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

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 12.1, Problem 59E

Chapter 12.1, Problem 61E

Students have asked these similar questions

Consider the graph ofr(t) = 2t^2i + t^2j + t^3 kDetermine parametric equations of the tangent line to the curve at the point where the curve intersects the planex − 2y − z = 8.

In Exercises 31-36, find a parametrization of the tangent line at the point indicated. (lns)i + s- 1 j + 9sk, s = 1

Find the unit tangent vector T(t) and a set of parametric equations for the tangent line r(t) = e2ti + cos tj − sin 3tk to the space curve at point P (1, 1, 0).

Chapter 12 Solutions

CALCULUS (CLOTH)

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

Knowledge Booster

$Calculus$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.

- Find an equation y = f ( x ) for the parametric curve and compute d y d x in two ways for c ( t ) = ( 1 2 t , 1 4 t 2 − t ) (a) Using d y d x = d y d t d x d t = y ' ( t ) x ' ( t ) . (b) By differentiation of f ( x ) .

Solution Summary: The author explains that for finding the equation y = f ( x ), we have to solve both parametric equations and eliminate the variable t.

Concept explainers

Videos

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

$c (t) = (\frac{1}{2} t, \frac{1}{4} t^{2} - t)$

(a) Using $\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{y' (t)}{x' (t)}$ .

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

Want to see the full answer?

Chapter 12 Solutions

- Find an equation y = f ( x ) for the parametric curve and compute d y d x in two ways for c ( t ) = ( 1 2 t , 1 4 t 2 − t ) (a) Using d y d x = d y d t d x d t = y ' ( t ) x ' ( t ) . (b) By differentiation of f ( x ) .

Solution Summary: The author explains that for finding the equation y = f ( x ), we have to solve both parametric equations and eliminate the variable t.

Concept explainers

Videos

- Find an equation y=f(x) for the parametric curve and compute dydx in two ways for c(t)=(12t,14t2−t) (a) Using dydx=dydtdxdt=y'(t)x'(t). (b) By differentiation of f(x).

Want to see the full answer?

Chapter 12 Solutions

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

$c (t) = (\frac{1}{2} t, \frac{1}{4} t^{2} - t)$

(a) Using $\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{y' (t)}{x' (t)}$ .

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