CALCU-Question 1Consider the curve with the rule 3

Solution by Steps

step 1

To find

\frac{dy}{dx}

using implicit differentiation, we start with the equation

3xy + x^2 = x

. Differentiating both sides with respect to

x

gives:

step 2

3y + 3x\frac{dy}{dx} + 2x = 1

. Now, we isolate

\frac{dy}{dx}

:

step 3

Rearranging the equation, we have

3x\frac{dy}{dx} = 1 - 3y - 2x

. Thus, we can express

\frac{dy}{dx}

as:

step 4

\frac{dy}{dx} = \frac{1 - 3y - 2x}{3x}

step 5

Now, to find

x

in terms of

y

when

\frac{dy}{dx} = \frac{1}{2}

, we set

\frac{1 - 3y - 2x}{3x} = \frac{1}{2}

step 6

Cross-multiplying gives

2(1 - 3y - 2x) = 3x

. Simplifying this leads to:

step 7

2 - 6y - 4x = 3x

. Rearranging gives

2 - 6y = 7x

. Thus, we can express

x

as:

step 8

x = \frac{2 - 6y}{7}

Answer

\frac{dy}{dx} = \frac{1 - 3y - 2x}{3x}

and

x = \frac{2 - 6y}{7}

Key Concept

Implicit differentiation allows us to find derivatives of functions defined by equations involving both

x

and

y

.

Explanation

The first part gives the derivative

\frac{dy}{dx}

in terms of

x

and

y

, while the second part provides

x

in terms of

y

when the slope is

\frac{1}{2}

.

Solution by Steps

step 1

To find the equation for the tangent to the curve with the gradient

\frac{1}{2}

when

y=5

, we first need to find the corresponding

x

value on the curve

step 2

We substitute

y=5

into the original curve equation

3xy + x^2 = x

to find

x

:

3x(5) + x^2 = x

step 3

This simplifies to

15x + x^2 - x = 0

or

x^2 + 14x = 0

. Factoring gives

x(x + 14) = 0

, so

x = 0

or

x = -14

step 4

We now have two points:

(0, 5)

and

(-14, 5)

. We will find the tangent line at both points using the gradient

\frac{1}{2}

step 5

The equation of the tangent line can be expressed as

y - y_1 = m(x - x_1)

, where

m = \frac{1}{2}

. For the point

(0, 5)

:

y - 5 = \frac{1}{2}(x - 0)

step 6

This simplifies to

y = \frac{1}{2}x + 5

. For the point

(-14, 5)

:

y - 5 = \frac{1}{2}(x + 14)

step 7

This simplifies to

y = \frac{1}{2}x + 12

. Thus, the equations of the tangents are

y = \frac{1}{2}x + 5

and

y = \frac{1}{2}x + 12

Answer

The equations of the tangents are

y = \frac{1}{2}x + 5

and

y = \frac{1}{2}x + 12

.

Key Concept

The tangent line to a curve at a given point represents the instantaneous rate of change of the function at that point.

Explanation

We found the points on the curve where

y=5

and used the given gradient to derive the equations of the tangent lines.

Solution by Steps

step 1

We start with the differential equation

\frac{dy}{dx} = x^2 - 2y

and the initial condition

y(2) = 1

. We need to find

y(2.1)

step 2

Using the initial condition, we substitute

x = 2

into the differential equation to find

\frac{dy}{dx}

at that point:

\frac{dy}{dx} = 2^2 - 2(1) = 4 - 2 = 2

step 3

We can use the approximation

y(2.1) \approx y(2) + \frac{dy}{dx} \cdot (2.1 - 2)

. Substituting the values gives

y(2.1) \approx 1 + 2 \cdot 0.1 = 1 + 0.2 = 1.2

Answer

y(2.1) \approx 1.2

Key Concept

Using differential equations to approximate values based on initial conditions

Explanation

We used the initial condition and the derivative at that point to estimate the function's value at a nearby point. This method is known as linear approximation.

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