Calculate dy/dx at the point t=10 for the equations:
a) x=5t
b) (t^2)-1
f(t)=c then f'(t)=0 i.e. a derivative of a constant (or number) is zero
f(t)=t^n then f'(t)=nt^(n-1)
you bring the exponent n in front and subtract one from the original exponent
finally, plug in 10 for t
a)50
b) 99
To calculate dy/dx at the point t = 10 for each equation, we need to find the derivatives of the equations with respect to t and then substitute t = 10 into the derivatives.
a) x = 5t:
To find dx/dt, we differentiate x = 5t with respect to t:
dx/dt = 5
Now, to find dy/dx, we need to use the chain rule, which states that dy/dx = dy/dt / dx/dt.
Since x = 5t, we can rearrange this equation to y = x:
dy/dt = dy/dx * dx/dt
dy/dt = 1 * 5 = 5
Finally, substituting t = 10 into dy/dt:
dy/dt = 5
Therefore, dy/dx at t = 10 for x = 5t is 5.
b) y = t^2 - 1:
To find dy/dt, we differentiate y = t^2 - 1 with respect to t:
dy/dt = 2t
Now, we need to calculate dx/dt. However, we don't have an equation for x in terms of t in this case. Since there is no direct relationship between x and t, we cannot find the derivative dx/dt.
Therefore, we cannot calculate dy/dx for y = t^2 - 1 at the point t = 10, as we don't have enough information about the relationship between x and t.