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3 years ago

Suppose xy=−4 and dy/dt=−3. Find dx/dt when x=−1.

1 answer:

4 0

When x=-1: $\quad (-1)y=-4\qquad\to\qquad y=4\)$

Ok that gives us a little more information.
If we implicitly differentiate with respect to t, from the very start, then we can apply our product rule, ya?

$x'y+xy'=0$

The right side is zero, derivative of a constant is zero.
Where x' is dx/dt and y' is dy/dt.

From here, plug in all the stuff you know:
y' = -3
x = -1
y = 4

and solve for x'.

Hope that helps!

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1. Find equation (both the normal equation and the parametric equation) for the plane (in R3) so that (a) the plane contains the

kherson [118]

Answer:

For a part:

(x-1, y, z-3) * (-6, -3, -2) = 0 (Normal equation)

P = (1,0,3) + (1,-2,0)*s + (0,-2,3)*t (Parametric equation)

For b part:

(x-1, y, z-3) * (1, 1, 1) = 0 (Normal equation)

P = (4, 0, 0) + (-1, 1, 0)*s + (-1, 0, 1)*t (Parametric equation)

Step-by-step explanation:

The parametric equation of a plane is defined by three things: a point, and two non-colinear vectors in the plane. We already have three points, so we only need to find two vectors contained in the plane (we will call them V1 and V2):

V1 = P-Q = (1,0,3)-(0,2,3) = (1,-2,0)

V2 = P-R = (1,0,3)-(1,2,0) = (0,-2,3)

Therefore, the parametric equation of our plane is given by:

P = P + V1*s + V2*t = (1,0,3) + (1,-2,0)*s + (0,-2,3)*t

The normal equation of any plane is given by [x-P]*n = 0, where x is the vector(x,y,z), P is a point contained in the plane, n is a vector normal to the plane and * stands for the dot product. Therefore, for finding the normal form of the equation, we need an orthogonal vector to the plane (n), which we find by doing the cross product of our previous vectors V1 and V2:

n = V1xV2 = (-6,-3,-2)

Substituting the required values in the formula mentioned above, we can write the normal equation of our plane as:

[(x,y,z) - (1,0,3)]*(-6,-3,-2) = 0 or (x-1, y, z-3) * (-6, -3, -2) = 0

In this second exercise, it is straightforward to give the normal equation:

[(x,y,z) - (1,0,3)]*(1,1,1) = 0 or (x-1, y, z-3) * (1, 1, 1) = 0

For transforming this equation to its parametric form, first we transform it to its cartesian form:

x-1 +y +z-3 = 0, then : x+y+z = 4

Now that we have the cartesian form, we solve for variable x and get:

x = 4 -y -z

Then, we know that every point in the plane can be expressed as:

P = (4 -y -z, y, z).

Finally, rewriting the expression and converting y and z to the parameters s and t respectively we get:

P = (4, 0, 0) + (-y, y, 0) + (-z, 0, z) = (4, 0, 0) + (-1, 1, 0)*s + (-1, 0, 1)*t

4 0

3 years ago

Which choice is equivalent to the quotient shown here for acceptable values of x?

OleMash [197]

Sqrt(5x) * Sqrt(x) + 3
= 5^(1/2) * x^(1/2) * x^(1/2) + 3
= 5^(1/2) *x +3
=5^(1/2)*x^(2/2) +3
=Sqrt(5*x^2) +2

So the answer is A

5 0

3 years ago

Read 2 more answers

2[3(4+ + 1] - 2 Please answer

Mariulka [41]

Answer:

Step-by-step explanation:

7 0

3 years ago

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EastWind [94]

Answer:

1) 4 bottles

2) $0.75

3) $5.25

Step-by-step explanation:

1)9/3 = 3 so you must divide 12 by 3

12/3=4 bottles

2)12/1=12

divide 9 by 12

9/12 =$0.75

3)12/7 = $\frac{12}{7}$ (The actually number is too long to write so just leave it in fraction form)

9÷12/7=$5.25

7 0

2 years ago

Brita has a monthly budget of x dollars. She spends $1,425 on her mortgage every month. One-third of her remaining budget is spe

bazaltina [42]

Answer:

$(x - 1425)/3

Step-by-step explanation:

Let total monthly budget be x

Amount spent on mortgage = $1425

Remaining balance after spending on mortgage = x - $1425

Amount spent on recreational activities = one-third of the balance

Expressing amount spent on recreational activities as an equation will give;

Amount spent on recreational activities = 1/3 of x - $1425

= 1/3 × ( x - $1425)

= ( x - $1425)/3

Hence the required equation is;

R = $(x - 1425)/3

R means recreational activities

3 0

2 years ago