Please, I need math help

Question

Hi, I really need help on these problems:

1. A point P moves along the curve in such a way that the y-coord is increasing at a constant rate of 0.06 units per second. Find the rate of change of the x-coord when P passes through (2,5).

2. A curve is such that dy/dx = sqrt(4x+1) and (2,5) is a point on the curve. Find the equation of the curve.

3. Show that (d^2y)/(dx^2) * (dy)/(dx) is constant.

4. Find the volume obtained when the shaded region is rotated through 360 degrees about the x-axis. Give you answer in terms of pi.

5. Part of a curve y=x/2 + 6/x is shown. The line y=4 intersects the curve at the points P and Q. Show the tangents to the curve at P and Q meet at a point on the line y=x.

Please help, I am quite clueless.

Answer 1

Hi, Marie. Iron Math is here to help!

Part 1: We are told that y increases at 0.06 units/second. This means that dy/dt = +0.06

To find dx/dt, we use the identity: dy/dx = (dy/dt) / (dx/dt) and then solve for dx/dt.

However, we need the equation of "the curve" to find dy/dx, the derivative of the curve's equation.

Without that equation, we can go no further. But it should be easy for you to find dy/dx and substitute above.

Part 2: Given dy/dx = sqrt(4x + 1)

To find y, we integrate this equation: y = integral (sqrt(4x + 1)) dx

= integral (4x + 1)^1/2 dx [always convert roots to powers - easier!]

Substitute u = 4x + 1. so du = 4 dx or dx = 1/4 du

So, y = (1/4) integral (u^1/2) du

= 1/4 u^3/2 / (3/2) + C

= 1/6 (4x + 1)^3/2 + C

To find the value of the integration constant, C, plug in the point on the curve, namely x =2, y =5

5 = 1/6 (4 x 2 + 1)^3/2 + C

30 = 9^3/2 + C [multiply both sides by 6]

30 = 27 + C [9^3/2 = (9^1/2)³ = 3³ = 27]

So, C = 3

The required equation is y = 1/6 (4x + 1) + 3

Part 3: Can't do this either, without knowing the specific equation!

Part 4: Same thing!

Part 5. Here, the curve is: y = x/2 + 6/x . . . . . . . . . . . . eqtn (1)

To find where the line y = 4 intersects the curve, just plug in the value y = 4.

That gives: 4 = x/2 + 6/x

4(2x) = x² + 12 [multiply through by 2x, the common denominator on the right]

Or: x² - 8x+ 12 = 0 which is a simple quadratic

Factor: (x-2)(x-6) = 0

Therefore, x = 2 or 6 [The horizontal line y = 4, intersects the curve in these 2 places]

Get the corresponding y values by plugging these values into eqtn (1) above

When x = 2, y = 2/2 + 6/2 = 4, so point P is (2, 4)

When x = 6, y = 6/2 + 6/6 = 4, and point Q is (6, 4)

Now, we need the tangents at P and Q.

Let the tangent lines by y = mx + b . . . . . . . . . .. . . . . . . . . . . . . . . . .. eqtn (2)

To find the slope(s), m, differentiate the equation: y = x/2 + 6/x

dy/dx = 1/2 - 6/x²

Hence, the slope at P = 1/2 - 6/2² = 1/2 - 1/4 = 1/4 [plug in x =2 to dy/dx]

And the slope at Q = 1/2 - 6/36 = 1/2 - 1/6 = 1/3

To find the b value in eqtn (2) for each tangent, we plug in the x and y values and slopes for P and Q

For P(2, 4): 4 =1/4 (2) + b => b = 7/2. So, the eqtn of the tangent at P is y = x/4 + 7/2

For Q(6, 4): 4 = 1/3 (6) + b => b = 2. And the eqtn of the tangent at Q is y = x/3 + 2

These 2 tangent lines meet when their x and y values are the same => x/4 + 7/2 = x/6 + 2

Solve for x and plug back into one of the equations.