# What is the instantaneous velocity of an object moving in accordance to # f(t)= (sin2t-cos2t,sin(2t-pi/4)) # at # t=(-pi)/3 #?

And makes an angle with the positive direction of

and its direction can be given through the relation

I hope this was helpful.

By signing up, you agree to our Terms of Service and Privacy Policy

To find the instantaneous velocity of an object moving according to the function ( f(t) = (\sin(2t) - \cos(2t), \sin(2t - \frac{\pi}{4})) ) at ( t = -\frac{\pi}{3} ), we first need to differentiate the function with respect to ( t ) to find the velocity vector function. Then, we substitute ( t = -\frac{\pi}{3} ) into the velocity vector function to find the instantaneous velocity at that particular time.

Taking the derivative of ( f(t) ) with respect to ( t ), we get:

[ \frac{df}{dt} = \left( \frac{d}{dt}(\sin(2t) - \cos(2t)), \frac{d}{dt}(\sin(2t - \frac{\pi}{4})) \right) ]

Using the chain rule and the derivatives of sine and cosine functions, we get:

[ \frac{df}{dt} = (2\cos(2t) + 2\sin(2t), 2\cos(2t - \frac{\pi}{4})) ]

Now, substitute ( t = -\frac{\pi}{3} ) into the velocity vector function:

[ \frac{df}{dt}\bigg|_{t = -\frac{\pi}{3}} = (2\cos(-\frac{2\pi}{3}) + 2\sin(-\frac{2\pi}{3}), 2\cos(-\frac{2\pi}{3} - \frac{\pi}{4})) ]

Simplify the trigonometric functions:

[ \frac{df}{dt}\bigg|_{t = -\frac{\pi}{3}} = (-\sqrt{3} - 2, 2\cos(-\frac{11\pi}{12})) ]

Finally, compute the cosine value:

[ \cos(-\frac{11\pi}{12}) \approx -0.2588 ]

Therefore, the instantaneous velocity of the object at ( t = -\frac{\pi}{3} ) is approximately:

[ (-\sqrt{3} - 2, 2 \times (-0.2588)) ]

[ \approx (-2.732, -0.5176) ]

By signing up, you agree to our Terms of Service and Privacy Policy

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- How do you use the definition of a derivative to find the derivative of #f(x) = |x|#?
- How do you find instantaneous velocity in calculus?
- How do you find the equation of the tangent line to the graph of #y=ln(x)# at x=1?
- What is the equation of the normal line of #f(x)=1/sqrt(x^2-2x+1)# at #x=2 #?
- What is the equation of the tangent line of #f(x)=(x^2+1)/(x+2)# at #x=-1#?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7