1)find line and the normal line:x^2+y^2=10 (3,1)

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1)find line and the normal line:x^2+y^2=10 (3,1)
2)find dy/dx:x^2+3xy+y^3=10
3)find dy/dx:y^2=(x-y)(x^2+y)
4)The temperature T(in degrees Fahrenheit)of food in a freezer is T=700/t^2+4t+10 where t is the time in hours.Find the rate of change of T with respect to t at each of the following times:t=1,t=3,t=5,t=10
希望可以把步骤写出来,

1、Your question 1 should ask for the tangent and the normal.
x² + y² = 10
2x + 2y*dy/dx = 0
dy/dx = -2x / 2y = -x/y
The slope at (3,1) is -3/1 = -3
So the tangent is y - 1 = -3(x - 3)
y = -3x + 10
The normal is y - 1 = -1/(-3) * (x-3)
y = (1/3)x
2、x² + 3xy + y³ = 10
2x + 3(y + x*dy/dx) + 3y²*dy/dx = 0
2x + 3y + 3x*dy/dx + 3y²*dy/dx = 0
dy/dx * (3x + 3y²) = -2x - 3y
dy/dx = -(2x+3y) / 3(x + y²)
3、y² = (x-y)(x²+y)
y² = x³-x²y+xy-y²
2y²+x²y-xy = x³
4y*dy/dx + (y*2x + x²*dy/dx) - (y + x*dy/dx) = 3x²
dy/dx * (4y + x² - x) = 3x² - 2xy + y
dy/dx = (3x²-2xy+y)/(x²-x+4y)
4、T = 700/(t² + 4t + 10)
dT/dt = 700 * -(2t + 4 + 0)/(t² + 4t + 10)² = -1400(t + 2)/(t² + 4t + 10)²
at t = 1,dT/dt = -1400(1+2)/(1+4+10)² = -56/3 ≈ -18.67 (°F/s)
at t = 3,dT/dt = -1400(3+2)/(9+12+10)² = -7000/961 ≈ -7.28 (°F/s)
at t = 5,dT/dt = -1400(5+2/)/(25+20+10)² = -392/121 ≈ -3.24 (°F/s)
at t = 10,dT/dt = -1400(10+2)/(100+40+10)² = -56/75 ≈ -0.75 (°F/s)

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