Consider the parametric curve given by the equations
x(t)=t^2–14t–4
y(t)=t^2–14t+48
How many units of distance are covered by the point P(t) = (x(t),y(t)) between t=0, and t=11 ?
distance, or displacement? There is a pronounced difference in those.
well to get a distance between points we use the distance formula but first we need the points themselves.
Point 1 is found by subbing t=0 in for t in x(t) and y(t)
Point 2 is found by subbing t=11 in for t
in x(t) and y(t)
I'll do point 1
x(t) = (0)^2 - 14(0) - 4 = -4
y(t) = (0)^2 - 14(0) + 48 = 48
so point 1 is (-4,48)
Hope that helps
To find the distance covered by the point P(t) between t=0 and t=11, we need to calculate the arc length of the parametric curve.
The arc length of a parametric curve given by (x(t), y(t)) between t=a and t=b can be calculated using the formula:
L = ∫[a,b] sqrt((x'(t))^2 + (y'(t))^2) dt
where x'(t) and y'(t) are the derivatives of x(t) and y(t) with respect to t.
Let's calculate the derivatives of x(t) and y(t) first:
x'(t) = 2t - 14
y'(t) = 2t - 14
Now, we can calculate the integrand of the arc length formula:
sqrt((x'(t))^2 + (y'(t))^2) = sqrt((2t - 14)^2 + (2t - 14)^2)
= sqrt(4t^2 - 56t + 196 + 4t^2 - 56t + 196)
= sqrt(8t^2 - 112t + 392)
= sqrt(8(t^2 - 14t + 49))
= sqrt(8(t - 7)^2)
Next, we can calculate the arc length using the integral:
L = ∫[0,11] sqrt(8(t - 7)^2) dt
= √8 ∫[0,11] |t - 7| dt
Since t - 7 is non-negative over the interval [0, 11], we can simplify the integral to:
L = √8 ∫[0,7] (t - 7) dt
= √8 [t^2/2 - 7t] evaluated from 0 to 7
= √8 [(7^2/2 - 7*7) - (0^2/2 - 7*0)]
= √8 [49/2 - 49 - 0]
= √8 (-49/2)
= -7√8
Therefore, the point P(t) covers a distance of 7√8 units between t=0 and t=11.
To find the distance covered by the point P(t) between t=0 and t=11, we need to calculate the arc length of the curve defined by the parametric equations x(t) and y(t).
The arc length formula for a parametric curve is given by:
L = ∫√[ (dx/dt)^2 + (dy/dt)^2 ] dt
In this case, we have x(t) = t^2–14t–4 and y(t) = t^2–14t+48. Let's find the derivatives of x(t) and y(t) to calculate the integrand in the formula.
- Differentiating x(t) with respect to t:
dx/dt = d/dt (t^2–14t–4)
= 2t–14
- Differentiating y(t) with respect to t:
dy/dt = d/dt (t^2–14t+48)
= 2t–14
Now, substitute these derivatives into the arc length formula:
L = ∫√[ (2t–14)^2 + (2t–14)^2 ] dt
= ∫√[ 4(t–7)^2 + 4(t–7)^2 ] dt
= 2 ∫√[ (t–7)^2 + (t–7)^2 ] dt
= 2 ∫2√[ (t–7)^2 ] dt
= 4 ∫|t–7| dt
To calculate the integral, we need to split it into two parts: t – 7 for t > 7, and 7 – t for t < 7. The factor of 4 can be taken out of the integral.
L = 4 ∫ (t–7) dt for t > 7
= 4 [(t^2/2) – 7t] + C
L = 4 ∫ (7–t) dt for t < 7
= 4 [7t – (t^2/2)] + C
To find the definite integral between t=0 and t=11, we need to evaluate these two integrals at these limits.
L = 4 [ (11^2/2) – 7(11) ] + 4 [7(11) – (11^2/2)] + C
= 4 [121/2 – 77] + 4 [77 – 121/2] + C
= 4 [121/2 – 77 + 77 – 121/2] + C
= 4 [0] + C
= 0 + C
= C
Therefore, the distance covered by the point P(t) between t=0 and t=11 is 0 units.