Explain how to do these questions~
1. An object moves in a straight line with its position at time t seconds given by s(t) = 3-2t+t^2, where s is the measured in metres. At what time is the object not moving?
2. Determine the coordinates of the point(s) on the graph of y=3x- 1/x at which the slope of the tangent is 7.
s=3-2t+t^2
ds/dt=-2+2t
when it is not moving, ds/dt =0, solve for t.
y=(3x-1)/x= 3-1/x
dy/dx=+1/x^2
if slope is 7, then
7=1/x^2
x^2=1/7
x= sqrt(1/7)
y=(3x-1)/x solve for y.
For number one we find the derivative of 3-2t+t^2
wouldn't that be t-2?
Oh crap nvm!
So for number 1 the answer is t=1
-2+2t=0
2t=2
2/2=1
1. To determine at what time the object is not moving, we need to find when its velocity is zero. Velocity is the derivative of position with respect to time, so we can find the velocity function by differentiating the position function, s(t).
First, let's find the derivative of s(t):
s'(t) = d/dt (3 - 2t + t^2)
= -2 + 2t
The object is not moving when its velocity is zero, so we need to solve the following equation:
-2 + 2t = 0
To solve for t, we isolate it by adding 2 to both sides:
2t = 2
Then, divide both sides by 2:
t = 1
Therefore, the object is not moving at t = 1 second.
2. To find the coordinates of the point(s) on the graph of y = 3x - 1/x where the slope of the tangent is 7, we need to find the derivative of the given function and then find the x-values where the derivative equals 7.
First, let's find the derivative dy/dx of y = 3x - 1/x using the power rule and the quotient rule:
dy/dx = d/dx (3x - 1/x)
= 3 - (-1/x^2)
= 3 + 1/x^2
= (3x^2 + 1)/x^2
Now, we need to find the x-values where the derivative is equal to 7:
(3x^2 + 1)/x^2 = 7
To simplify the equation, we can multiply both sides by x^2:
3x^2 + 1 = 7x^2
Next, we rearrange the equation:
7x^2 - 3x^2 = 1
Combine like terms:
4x^2 = 1
To isolate x^2, divide both sides by 4:
x^2 = 1/4
Take the square root of both sides:
x = ±√(1/4)
Simplifying further, we get two possible x-values:
x = ±1/2
Now that we have the x-values, we can substitute them back into the original equation to find the corresponding y-values.
When x = 1/2:
y = 3(1/2) - 1/(1/2)
= 3/2 - 2
= -1/2
So, one of the points where the slope of the tangent is 7 is (1/2, -1/2).
When x = -1/2:
y = 3(-1/2) - 1/(-1/2)
= -3/2 + 2
= 1/2
The other point where the slope of the tangent is 7 is (-1/2, 1/2).