Am I Right? (Differentiation)

**1. Find the equation of the tangent at the given points for: f(x) = x**^{2} + 3x, (2, 10)

I did:

f'(x) = 2x + 3

Let x = 2

--> m = 7

y - 10 = 7(x - 2)

y = 7x - 4

**2. Find the equation of the tangent to the curve y = x**^{3} - 9x^{2} + 20x - 8 at the point (1, 4). At what points of the curve is the tangent parallel to the line y = -4x + 3?

First, I used the method from the previous question to obtain the equation for the parallel line. For the next part of the question, I equated the gradient function to the gradient of the parallel line and solved for x, substituting the newly found x values into the original curve equation to get my y-coordinates.

**3. A model plane flying level at 250 m above the ground suddenly dives... blahablahhal... h(t) = 8t**^{2} - 80t + 250. Find the rate at which the plane is losing height at t = 3.

I simply substituted t = 3 into the original equation and solved, giving my answer as 82 ms^{-1}. Is this question as simple as that?

**4. A car starts from rest and moves a distance s (m) in t (s), where s = (1/6)t**^{3} + (1/4)t^{2}. What is the acceleration when t = 2?

I know that velocity = ds/dt and acceleration = dv/dt, so I found that s' = (1/2)t^{2} + (1/2)t and v' = t + (1/2). Then, to answer the question, I simply substituted t = 2 into the equation for acceleration and solved, giving my (5/2) ms^{-2}.

Thanks in advance. Your feedback is greatly appreciated.

Re: Am I Right? (Differentiation)

Quote:

Originally Posted by

**Fratricide** **1. Find the equation of the tangent at the given points for: f(x) = x**^{2} + 3x, (2, 10)

I did:

f'(x) = 2x + 3

Let x = 2

--> m = 7

y - 10 = 7(x - 2)

y = 7x - 4

Yes,, that is correct.

Quote:

**2. Find the equation of the tangent to the curve y = x**^{3} - 9x^{2} + 20x - 8 at the point (1, 4). At what points of the curve is the tangent parallel to the line y = -4x + 3?

First, I used the method from the previous question to obtain the equation for the parallel line. For the next part of the question, I equated the gradient function to the gradient of the parallel line and solved for x, substituting the newly found x values into the original curve equation to get my y-coordinates.

Sounds good. Why have you not done it?

Quote:

**3. A model plane flying level at 250 m above the ground suddenly dives... blahablahhal... h(t) = 8t**^{2} - 80t + 250. Find the rate at which the plane is losing height at t = 3.

I simply substituted t = 3 into the original equation and solved, giving my answer as 82 ms^{-1}. Is this question as simple as that?

No, it's not. Putting h= 3 into h(t) gives h(3)= 82 meters- not "meters per second", because, as you were told, h(t) is the **height** of the plane at t seconds, NOT the rate at which that height is changing.

Quote:

**4. A car starts from rest and moves a distance s (m) in t (s), where s = (1/6)t**^{3} + (1/4)t^{2}. What is the acceleration when t = 2?

I know that velocity = ds/dt and acceleration = dv/dt, so I found that s' = (1/2)t^{2} + (1/2)t and v' = t + (1/2). Then, to answer the question, I simply substituted t = 2 into the equation for acceleration and solved, giving my (5/2) ms^{-2}.

Yes, that is correct. Why didn't you do that for the previous question?

Quote:

Thanks in advance. Your feedback is greatly appreciated.

Re: Am I Right? (Differentiation)

Quote:

Originally Posted by

**HallsofIvy** Sounds good. Why have you not done it?

I have, I just didn't include it for brevity's sake.

Here it is:

dy/dx = 3x^{2} - 18x + 20

Let x = 1

--> m = 5

y - 4 = 5(x - 1)

y = 5x - 1

y = -4x + 3

m = -4

Let 3x^{2} - 18x + 20 = -4

3x^{2} - 18x + 24 = 0

(3x - 12)(x - 2) = 0

x = 2, 4

Let x = 2

y = 2(2)^{3} - 9(2)^{2} + 20(2) - 8

y = 4

Let x = 4

y = 2(4)^{3} - 9(4)^{2} + 20(4) - 8

y = -8

--> Tangent is parallel to the line y = -4x + 3 at points (2,4) and (4,-8).

Quote:

Originally Posted by

**HallsofIvy** Why didn't you do that for the previous question?

Ahhhh, I see. So I need to use the same method as in 4, substituting t = 3 into v'?

Re: Am I Right? (Differentiation)

Ignore that last part of my previous post. I'm pretty sure that for question three I need to find the velocity (not acceleration) at the given time.

As such, I did:

h'(t) = 16t - 80

Let t = 3

h'(t) = -32 ms^{-1}

Am I on the right track? (With this question and the workings out in my previous post.)

Re: Am I Right? (Differentiation)

I'm pretty sure what I've done is correct, but can anyone confirm it for me?

Re: Am I Right? (Differentiation)

If you get h'(3) = -32, then the plane is descending at 32m/s.