Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.
x=e^t ,y=te^t ,z=te^(t^2) ; (1,0,0)

Answers

Answer 1

The parametric equation of a tangent line is x(t) = 1+t, y(t) = t, z(t) = 1.

What is the parametric equation?

A parametric equation is a sort of equation that uses an independent variable known as a parameter (commonly indicated by t) and in which dependent variables are expressed as continuous functions of the parameter and are not reliant on another variable.

Here, we have

Given: x = [tex]e^{t}[/tex] ,y = t[tex]e^{t}[/tex] ,z = t[tex]e^{t^2}[/tex] ; (1,0,0)

We have to find the parametric equations for the tangent line to the curve.

r(t) = <  [tex]e^{t}[/tex] , t[tex]e^{t}[/tex] , t[tex]e^{t^2}[/tex]>

For, t = 0

r(0) = <1, 0, 0>

Now, we differentiate r(t) with respect to t and we get

r'(t) = < [tex]e^{t}[/tex], [tex]e^{t} +te^{t}[/tex], [tex]e^{t^2}+2t^2 e^{t^2}[/tex]>

At (1,0,0) , t = 0

r'(t) = < 1, 1, 1>

The equation of tangent line is given by:

<x(t),y(t),z(t)> =<1,0,0> + <1,1,1>t

= <1,0,0> + <t,t,t>

= <1+t,t,t>

Hence, the parametric equation of a tangent line is x(t) = 1+t, y(t) = t, z(t) = 1.

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Related Questions

Which statement is correct about the total number of functions from {a,b,c; to {1,21?
(A) The total number of functions from (1,2) to {a,b,c) is 9, and the number that are onto is 6.
(B) The total number of functions from (1,2) to {a,b,c) is 8, and the number that are onto is 6.
(C) The total number of functions from (1,2} to (a,b,c} is 9, and the number that are onto is 4.
(D) The total number of functions from {1,2) to {a,b,c) is 8, and the number that are onto is 4.

Answers

the correct statement about the total number of functions from {a,b,c; to {1,21 is (D) The total number of functions from {1,2) to {a,b,c) is 8, and the number that are onto is 4.

The total number of functions from {a, b, c} to {1, 2} is calculated by multiplying the cardinalities of the two sets.

Hence, the total number of functions is [tex]2^3 = 8[/tex](since there are three elements in the set {a, b, c} and two elements in the set {1, 2}).

Onto Function: A function f from set A to set B is called onto function if every element of B is the image of some element of A, which means that every element of B is a function of A.

We are asked to find the number of onto functions between these sets.

We know that if |A| < |B|, then there are no onto functions from A to B.

Here, |A| = 3 and |B| = 2. So, there cannot be an onto function from A to B.

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0/8 pts 499 Details Let y = 4x? Round your answers to four decimals if necessary. (a) Find the change in y, Ay when a I 7 and Ar = 0.2 Δy = (b) Find the differential dy when I = 7 and da = 0.2 Questi

Answers

For the equation y = 4x, the change in y, Δy, when x changes by 0.2 is 0.8. The differential dy, representing the instantaneous change in y when x changes by 0.2, is also 0.8.

(a) To find the change in y, denoted as Δy, when x changes by Δx, we can use the equation Δy = 4Δx. Since in this case Δx = 0.2, we can substitute the values to find Δy.

Δy = 4 * 0.2 = 0.8

Therefore, the change in y, Δy, is 0.8.

(b) The differential dy represents the instantaneous change in y, denoted as dy, when x changes by dx. In this case, dx is given as 0.2. We can use the derivative of y with respect to x, which is dy/dx = 4, to find the differential dy.

dy = (dy/dx) * dx = 4 * 0.2 = 0.8

Therefore, the differential dy is 0.8.

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Find two sets of parametric equations for the rectangular equation y = 32-2 1.2 t and y= 2. ytand =

Answers

The parametric equations for the rectangular equation y = 32 - 2(1.2t) are: x = t  y = 32 - 2(1.2t)  the second set of parametric equations is: x = 2t

y = y.

To find two sets of parametric equations for the rectangular equation y = 32 - 2(1.2t) and y = 2y_tan(t), we can assign different variables to represent x and y, and then express x and y in terms of those variables.

First set of parametric equations:

Let's use x = t and y = 32 - 2(1.2t).

x = t

y = 32 - 2(1.2t)

The parametric equations for the rectangular equation y = 32 - 2(1.2t) are:

x = t

y = 32 - 2(1.2t)

Second set of parametric equations:

Let's use x = 2t and y = 2y_tan(t).

x = 2t

y = 2y_tan(t)

To express y_tan(t) in terms of x and y, we can divide both sides of the second equation by 2:

y_tan(t) = y/2

The parametric equations for the rectangular equation y = 2y_tan(t) are:

x = 2t

y = 2(y/2) = y

Therefore, the second set of parametric equations is:

x = 2t

y = y

Note: In the second set of parametric equations, y is not explicitly defined in terms of x, as the equation y = y implies that the value of y remains constant throughout.

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Let . Then lim h-0 f(x+h)-f(x) h f(x) = x² - 2x + 7. 8.

Answers

To find the limit of the expression (f(x+h)-f(x))/h as h approaches 0, where f(x) = x² - 2x + 7, we can directly substitute the given function into the expression and simplify to obtain the limit.

The given function is f(x) = x² - 2x + 7. We are interested in finding the limit of the expression (f(x+h)-f(x))/h as h approaches 0. Let's substitute the function into the expression:

lim(h->0) (f(x+h)-f(x))/h = lim(h->0) ((x+h)² - 2(x+h) + 7 - (x² - 2x + 7))/h

Simplifying further:

= lim(h->0) (x² + 2xh + h² - 2x - 2h + 7 - x² + 2x - 7)/h

= lim(h->0) (2xh + h² - 2h)/h

= lim(h->0) 2x + h - 2

Since h is approaching 0, the term h will disappear, and we are left with:

= 2x - 2

Therefore, the limit of the expression (f(x+h)-f(x))/h as h approaches 0 is 2x - 2.

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Find the flux of the vector field F= (-yx.1) across the cylinder y = 5x?, for OsXs2,0528 1. Normal vectors point in the general direction of the positive y-axis. Parametrize the surface using u=x and

Answers

The flux of the vector field F across the cylinder y = 5x is 0. This means that the net flow of the vector field through the surface of the cylinder is zero.

To find the flux of the vector field F across the given cylinder, we need to calculate the surface integral of F over the surface of the cylinder. The surface of the cylinder can be parametrized using u = x and v = y. The normal vector to the surface of the cylinder points in the general direction of the positive y-axis.

Since the vector field F = (-yx, 1, 0), we can compute the dot product of F with the unit normal vector to the surface of the cylinder. The dot product represents the component of the vector field that is normal to the surface. However, since the normal vector and the vector field are perpendicular to each other, the dot product evaluates to zero. This implies that there is no net flow of the vector field through the surface of the cylinder.

In conclusion, the flux of the vector field F across the cylinder y = 5x is zero, indicating that there is no net flow of the vector field through the surface of the cylinder.

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